WBBSE Solutions For Class 10 Maths Statistics Chapter 2 Ogive

Statistics Chapter 2 Ogive

Ogive is the curved line or graph of a cumulative frequency distribution (Less or more than type).

By the help of ogive, we can find the median of a frequency distribution. There are two types of ogives

  1. Ogive of less than type cumulative frequency distribution
  2. Ogive of more than type cumulative frequency distribution.

Construction of Ogive:

We can construct two types of giving

  1. Ogive of less than type cumulative frequency distribution;
  2. Ogive of more than type cumulative frequency distribution.

Construction of ogive of less than type cumulative frequency distribution:

In order to construct the ogive of less than type cumulative frequency distribution, we first, determine the scale of the graph paper, i.e., we fixed up a certain scale of both the axes of the graph paper.

WBBSE Solutions for Class 10 Maths

For cumulative frequency, we denote the upper boundaries of each class interval along the horizontal line, i.e. along the x-axis, and their corresponding cumulative frequency along the- vertical line, i.e., along the y-axis.

Later on, the points thus obtained are plotted in the graph paper. Now, the points are joined by a scale or in open hands to get a curved
line.

This curved line is called ogive. Observe the following examples:

Example:

 

Class intervals 100-20 120-140 140-160 160-180 180-200
Frequencies 12 14 8 6 10

 

Find the median of the above frequency distribution by constructing ogives of less than and more than type cumulative frequencies along the same axes.

Solution: At first, let us construct a less-than-type and a more-than-type cumulative frequency distribution table of the above data.

 

Class interval Classes More than type cumulative frequencies Classes Less than type cumulative frequencies
100-120 100 or more than 100 50 Less than  120 12
120-140 120 or more than 120 38 Less than  140 26
140-160 140 or more than 140 24 Less than  160 34
160-180 160 or more than 160 16 Less than  180 40
180-200 180 or more than 180 10 Less than  200 50

 

Selection of Scale:

Let one side of each smallest square = 10 units along the x-axis and one side of each smallest square = 2 units along the y-axis.

Plotting of points:

For ogive of more than type cumulative frequency distribution, let us plot the points ( 1 00, 50), (120,38), (140, 24), (160, 16), and (180, 10).

But according to the above scale (10, 25), (12, 19), (14, 12), (16, 8) and (1 8, 5). Later on, let us plot the points on the same graph paper.

Then, joining the points by free-hand of by a scale, we get an ogive of more than type cumulative frequency.

For ogive of less than type cumulative frequency distribution, let us plot the points ( 1 20, 1 2), (140, 26),(160, 34), (180, 40), and (200, 50).

But according to the above scale the points will be transferred to the new points (12, 6), (14, 13), (16, 17), (18, 20), and (20, 25). Then let us plot the points on the same graph paper.

We thus get an Ogive of less than type cumulative frequency distribution.

We see from the graph paper that the two ogives intersect each other at P.

Now, let us draw a perpendicular PM from the point P on the x-axis. Taking measure of the perpendicular PM by a scale we get, PM = 13.857 (approx.)

But according to the scale, 13.857 square = 13.857 x 10 units (approx.) = 138.57 units (approx.)

The median of the data = 138-57 units (approx.)

We can thus determine the median of any given data by constructing ogives.

Now, we prove this determination of this median by formula directly:

Formula: Median = \(l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h\)

 

WBBSE Solutions For Class 10 Maths Statistics Chapter 2 Ogive

 

Here, 10 = 140, n = 50, \(\frac{n}{2}\) = 25, cf = 26, f = 14, h = 20

∴ Median = 140 + \(\frac{25-26}{14}\) x 20 = 140 – 1.4285 = 138.57

∴ The required median = 138.57 units(approx).

 

 

 

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode

Statistics Chapter 3 Median And Mode

The median or the mid-values is another measure of central tendency.

If the values of the variable x be arranged in ascending or descending order, then the value in the middle position of the arrangement is called the median.

There are equal number of values above and after the median.

Let x1,x2,……,xbe n-values of the variable x. If we arrange the values in ascending order either, we get, \(x_{(1)}, x_{(2)}, \ldots \ldots, x_{(n)}, \text { i.e., } x_{(1)} \leq x_{(2)} \ldots \ldots \leq x_{(n)}\)

Amongst the values, there may occur some values which are equal to each other. But not all are equal.

Now, if n is an odd number, then the \(\left(\frac{n+1}{2}\right)\)-th value is the median or mid-value.

WBBSE Solutions for Class 10 Maths

∴ the median = \(\mathrm{Me}=x_{\left(\frac{n+1}{2}\right)}\)…….(1)

If n is an even number, then we shall get no unique median of the values of x.

Then the two values \(\left(\frac{n}{2}\right)-\text { th and }\left(\frac{n}{2}+1\right)\)-th values are denoted as the medians of the values i.e,

here, \(x_{\left(\frac{n}{2}\right)} \leq \mathrm{Me} \leq x_{\left(\frac{n}{2}+1\right)}\)……(2)

For unique median we generally find the average of the two values \(\left(\frac{n}{2}\right)-\text { th and }\left(\frac{n}{2}+1\right)\)-th values, i.e.,

\(\mathrm{Me}=\frac{x_{\left(\frac{n}{2}\right)}+x_{\left(\frac{n}{2}+1\right)}}{2}\)

 

Statistics Chapter 3 Median And Mode Examples

Example 1. Find the median of the values in 15, 19, and 20.

Solution:

Given:

15, 19, and 20

Here, arranging the values in ascending order, we get, 15, 19, 20.

Here, the number of numbers is 3.

∴ the \(\left(\frac{3+1}{2}\right)\)-th i.e. the second value is the required median.

∴ Median = 19 quintals.

Example 2. Find the median of the values 17, 25, 30, 32, 28, 24, 20, 18, 16, 10 Arranging the values in ascending order we get 10, 16, 17, 18, 20, 24, 25, 28, 30, 32.

Solution:

Given:

10, 16, 17, 18, 20, 24, 25, 28, 30, 32

Here, the number of numbers is 10, which is an even number.

So, there are two mid-values, the \(\left(\frac{n}{2}\right) \text {-th and }\left(\frac{n}{2}+1\right)\)-th values, i,e., the 5th and 6th values are mid-values.

Thus, the two mid-values are 20 and 24.

∴ the required median = \(\frac{20+24}{2}\) = 22

∴ the required median = 22 kilometers/hour.

Example 3. Find the median of the values are given in below table

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 3

Solution: Here, arranging the values in ascending order, we get,

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 3-1

Here, the number of values is 41 and so the median is the \(\left(\frac{41+1}{2}\right)\)th i.e- 21st value.

Here, the 21st value is 24.

∴ the required median = 24.

Example 4. Find the median of clothes 2 shares of prices Rs.30 each, 3 sharees of process Rs.35 each, and 5 sharees of price Rs.45 each is given.

Solution:

Given:

clothes 2 shares of prices Rs.30 each, 3 sharees of process Rs.35 each, and 5 sharees of price Rs.45 each is given.

Here, amongst 10 clothes, the price of 2 is Rs. 30, the price of 3 is Rs. 35 and the price of 5 is Rs. 45.

Now, arranging the prices in ascending order we get 30,30,35,3.5,35,45,45,45,45,45.

Here, the total number of clothes is 10 which is an even number, the mid-values are 2 in number, i.e. the \(\left(\frac{10}{2}\right)\)th and \(\left(\frac{10}{2}+1\right)\)th values are the mid-values, i.e., the 5th and 6th values are the mid-values.

Here the 5th value = 35 and the 6th value = 45.

∴ the median = \(\frac{35+45}{2}\) = 40.

∴ the required median = Rs. 40.

Example 5. Find the median of the telephone calls given in the below table

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 5

Solution: Here, the values of the frequency distributions are already arranged in ascending order.

In this case, 0 occurs 2 times, and 1 occurs 5 times. 2 occurs 6 times, etc.

Now, preparing the cumulative frequency distribution table, we get,

Table: Finding of the median of the telephone calls

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 5-1

From the table, we see that the number of values up to 3 is 24, and up to 4 it is 36.

Here, the total number of values is 60.

So, the average of the 30th and 31st values is to be taken as the median.

Since every value from the 25th to 36th values is 4, so both the 30th and 31st values is 4. ∴ the median = 4.

Example 6. Find the medians of the diameters Here, also we have to prepare the following cumulative frequency distribution.

Solution:

Determination of the median of the diameters

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 6

Here, 25 values are each 45, 28 values are each 46, etc.

From the cumulative frequency, we see that the number of values up to 46 is 53, up to 47 is 85.

Here, the total number of values is 150. So, the 75th and 76th values are to be taken as the median.

Since from 54th to 85th values are all.47, so both 75th and 76th values are 47.

∴ the required median = 47 mm

Example 7. Find the median of heights in the given table.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 7

Solution: For continuous variables, at first we have to find the class boundaries and their corresponding cumulative frequencies.

The value for which the cumulative frequency is \(\frac{n}{2}\) is assumed to be the median.

Since here the frequencies of each class interval are given and the frequencies of different values are not available, it is impossible to determine the correct value of the median.

Frequencies of the different class intervals are equally distributed assuming this principle, we can find the required median. To do so, let us prepare the following tables:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 7-1

Here, \(\frac{n}{2}\) = \(\frac{100}{2}\) = 50.

From the cumulative frequencies we see that up to 159.5, the total frequency is 33, and up to 164.5, it is 63.

The value 50 lies in between 33 and 63. So, the median will lie in between the class boundary 159.5 to 164.5.

The frequency of the two class boundaries is 159.5 and 164.5.

The frequency of the two class boundaries 159.5 and 164.5 is 30 and the class length is 5.

So, by the method of proportional part, we get,

\(\frac{\text { Median }- \text { Lower class boundary }}{\text { Upper class boundary – Lower class boundary }}\)=\(\frac{50-33}{63-33}\)

 

or, \(\frac{\text { Median }-159 \cdot 5}{164 \cdot 5-159 \cdot 5}=\frac{50-33}{63-33} \text { or, Median }-159 \cdot 5=\frac{17}{30} \times 5\)

or, Median – 159.5 = 2.83 or, Median = (159.5 + 2.83) cm = 162.33 cm

In general, if xi be the lower boundaries of the class-interval jn which the median belongs to f0 be the frequency of that class interval F be the cumulative frequency upto xi, n be the total frequency, and c be the class length, then

Median, \(\mathrm{Me}=x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c\)……..(4)

Otherwise, for the value of \(\frac{n}{2}\), the value of the x-coordinate of the ogive (less or greater than type) is taken as the median.

Example 8. Find the median of the numbers frequency distribution of the marks obtained by 76 students.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 8

Solution: We have to prepare the following table:

Determination of the median of the numbers:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 8-1

Here, total frequency = 76 and \(\frac{n}{2}\) = 38.

38 is in between 18 and 50. So, the median lie iu between 20.50 and 30.5.

Here x1= 20.5, F = 18, f0 = 32, c =10

∴ Median = \(x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c=20 \cdot 5+\frac{38-18}{32} \times 10\)

= \(20 \cdot 5+\frac{20}{32} \times 10=20 \cdot 5+6 \cdot 25=26 \cdot 75\)

∴ the required median = 26.75

Example 9. The frequency distribution of the incomes of 75 employees of the industry are given below. Find the median.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 9

Solution: Determination of the median of the incomes :

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Example 9-1

Here, the total frequency n = 75 and \(\frac{n}{2}\) = 37.5 lies in between 20 and 45.

So, the median lies in the class interval 250- 500.

Here, x1 = 250, F = 20, f0 = 25, c = 250.

∴ Median = \(250+\frac{37 \cdot 5-20}{25} \times 250=250+\frac{17 \cdot 5}{25} \times 250=250+175=425\)

∴ the required median = Rs 425.

[Note: In the above example the first and last class intervals are open class-interval.]

Mode:

Mode is another measure of the central tendency. The value of the variable x, which occurs the highest times than all other values of the variable x in the given data is known as the mode of that data.

We briefly denote it by MQ.

For discrete variables, the greatest frequency of a frequency distribution is usually called its mode.

For a continuous variable, the value which have the highest frequency density is known as the mode.

Sometimes mere may occur values more than once, which have, the highest frequency or frequency density.

In these cases, it is impossible to determine the mode uniquely. Here, the variable or the given frequency distribution is said to be of several modes.

 

Statistics Chapter 3 Median And Mode Mode Examples

Example 1. Find the mode of the data is 15 quintals, 19 quintals, and 20 quintals are given.

Solution:

Given:

Data is 15 quintals, 19 quintals, and 20 quintals .

Here, the three given Values of the variable are 15 quintals, 19 quintals, and 20 quintals, which occur once each in the data.

So, here it is impossible, to determine mode uniquely.

Example 2. Find the mode of velocities 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 are given.

Solution:

Given:

velocities 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10.

Here also, each value occurs once in the given data.

So, here also it is impossible to determine the mode uniquely.

Example 3. Find the mode of the numbers are given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 3

Solution: Observe that each of the numbers 17, 21, 23, and 25 occur in 3 times each.

So, here also it is impossible to determine the mode uniquely.

Example 4. Find the mode of the selling prices 115, 98, 102, 126, 85, 91, and 107 are given.

Solution:

Given:

The selling prices 115, 98, 102, 126, 85, 91, and 107

Here, the 7 different values each occurs once in the given data.

So, here also it is impossible to find the mode uniquely.

Example 5. Find the mode of the prices of clothes 2 shares of price Rs.30 each, 3 shares of price Rs.35 each, and 5 shares of price Rs. 45 each.

Solution: Here, Rs.30 occurs 2 times, Rs. 35 occurs 3 times, and Rs. 45 occurs 5 times, i.e., Rs. 45 occurs the highest times.

So, the required mode = Rs 45.

Example 6. Find the mode of the telephone calls given per minute for any hour are given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 6

Solution: Here, telephone calls are discrete variable.

We can easily find from the frequency distribution that the frequency of 4 is 12 and it is the highest time to be occurred.

∴ The mode = 4.

Example 7. Find the mode of the frequency distribution of diameters.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 7

Solution: Here, amongst the 5 different values of the variable, the frequency of 4 mm is 35 and it is the highest.

∴The required mode = 48 mm

Example 8. Find the mode of the frequency distribution of the height (in cm) of 100 students.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 8

Solution: Here, height is a continuous variable.

It is very difficult to determine the mode of a continuous variable.

The easiest way, to determine the mode of a continuous variable, is to construct a histogram of the variable and then a similar or congruent frequency line along with the histogram.

The value of the variable for which the frequency line is the highest is its mode.

In the following, here is an imaginary histogram and its frequency line and so that the mode is determined.

 

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Of The Height

 

To determine this frequency line congruently is not an easy matter.

Roughly we may take the mid-value of the class interval which have the highest frequency density as the mode.

i.e., \(\mathrm{M}_0 \cong \frac{x_u+x_l}{2}=x_0 \text { (let) }\)…..(5)

Again, let the class intervals are equal and each of them be c.

If the highest frequency be f0, the frequency of class interval of the previous class interval be f(-1) and the frequency of the class interval of the next class interval be f1.

Now, it is possible to take x0 as the mode if f0– f(-1) and f0– f1 are equal.

If f0– f(-1) is greater than f0– f1, the mode is nearer to xi and if it is alternative, the mode is nearer to x1.

According to the proportional law,

\(\frac{\mathrm{M}_0-x_l}{x_u-\mathrm{M}_0}=\frac{f_0-f_{-1}}{f_0-f_1}\)

 

or, \(\frac{\mathrm{M}_0-x_l}{x_u-x_l}=\frac{f_0-f_{-1}}{2 f_0-f_1-f_{-1}}\)

or, \(\mathrm{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_1-f_{-1}} \times c \quad\left[because x_u-x_l=c\right]\)

\( \bar{x}-M_0=3\left(\bar{x}-M_e\right)\)

 

or, \(\mathrm{M}_0=3 \mathrm{M}_e-2 \bar{x}\)…….(6)

There is no proof of the above relation, but in maximum cases, it is seen that the relation (6) is roughly satisfied.

Using the formula (4), we get

\(\text { Mode }=\mathrm{M}_0=159 \cdot 5+\frac{30-22}{2 \times 30-22-24} \times 5, \text { where } \mathrm{f}_0=30, \mathrm{f}_{-1}=22, \mathrm{f}_1=24, \mathrm{c}=5\)

 

and \(\mathrm{x}_1=159 \cdot 5 \text { and } \mathrm{x}_{\mathrm{u}}\)

= 164.5

= \(\left(159 \cdot 5+\frac{8}{14} \times 5\right) \mathrm{cm}\) =(159.5 + 2.86)= 162.36 cm

Again, from the formula (6), we get,

Mode = \(\mathrm{M}_0=3 \times \mathrm{Me}-2 \bar{x}=3 \times 162 \cdot 33-2 \times 162 \cdot 40,\)

[where \(\bar{x}\) =162.40 and \(\mathrm{M}_e=162 \cdot 33\)]

Also, from the formula (6), we get,

Mode = 3 x Me – 2\(\bar{x}\)

= 3 x 162.33 – 2 x 162 = 486.99 – 324.80 = 162.19

∴ the required mode = 162.19 cm.

Example 9. Find the mode of the marks frequency distribution obtained by 76 students.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 9

Solution: Here, the mode lies in the interval 20.5- 30.5, since its frequency density is the highest.

∴ x1= 20.5, xu = 30.5, c = 10, f0 = 32, f-1 = 14 and f1 = 18

∴ By formula, the mode = 20.5 + \(\frac{32-14}{2 \times 32-14-18}\) x 10 = 20.5 + \(\frac{18}{32}\) x 10

= 20.5 + 5.62 = 26.12 (approx.)

By another formula, Me = 26.75 and \(\bar{x}\) = 27.08,

Mode = 3 x Me– 2\(\bar{x}\) = 3 x 26.75 – 2 x 27.08 = 80.25 – 54.16 = 26.09

Comparison of Mean, Median, and Mode

Mean, median add mode all three are measures of central tendency.

All three measures are easy, comprehensible, and have comprehensible significance.

But to the general public, Mean is very well known and is widely used.

All three measures can easily be determined, although for the continuous variable, it is difficult to determine to die mode and we generally become satisfied by its approximate value Among the three measurements, only the mean is dependable directly on all the obtained values of the variable because in the case of other two measurements, we can change the other values of the measurements by taking the measurement unchangeable.

If the mid-values and mode remain unchanged, then the median and mode also remain unchanged.

However, you have seen that the Mean has some easy properties, so that is it easy to use. Median and mode have no such properties.

There are some special areas where the Use of Mean is impossible and improper. Let amongst 7 person income of 6 persons are in between Rs. 500 and Rs. 600 and the income of the 7th person is Rs. 2000.

If we determine the Mean of it we shall see that it is more than Rs.-600 and it can never be the measure of central tendency.

So, if some of the values be more greater or smaller compared to the other values, then it is improper to determine the mean of the variable.

Here the first and last class intervals are open. As a result, determining the mid-values of that two class intervals are impossible. So, it is also impossible to find out the mean.

In this case, it is easier to find out the median. So if the first and last or both the class intervals are open, then it is reasonable to determine the median and mode as the central tendency.

Example 10. The median and mode of the 61 values of a variable are 17 and 19 respectively. Later on, it is seen that one of the values is wrongly taken as 13, the correct value of which is 16. Find the correct values of median and mode.

Solution:

Given:

The median and mode of the 61 values of a variable are 17 and 19 respectively. Later on, it is seen that one of the values is wrongly taken as 13, the correct value of which is 16.

If we arrange 61 values of the variable in ascending or descending order, then by definition, the 1st i.e., the 31st value is the median. The wrong number is 13.

So, if we arrange the values in ascending order, then its order of value will be below 31. Again, instead of 13, if 16 is taken, then the value of the mean will be also below the 31st value.

So, the median = 31st value will remain the same. Mode is the value of the variable with the highest frequency. Therefore, if any other values of the variable be changed, then the mode will remain unchanged.

So, the mode remains 19.

Example 11. The mean of the frequency distribution of a variable is 25 and its median is 23. Find the value of its mode.

Solution:

Given:

The mean of the frequency distribution of a variable is 25 and its median is 23.

We know that the relationship between mean, median, and mode is given by Mean- Mode = 3 (Mean – Median)

∴ Mode = 3 x Median- 2 x Mean = 3 x 23- 2 x 25 = 69- 50 = 19

∴ the required mode =19.

Example 12. Find the Median and Mode of the variable from the following frequency distribution table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 12

Solution: Here, the class ranges are not equal. There will be no problem in determining the median, but to find the mode, the class ranges must be made equal.

Let us draw the following table of cumulative frequency distribution for the median.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 12-1

We know that median, \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{120}{2}\) = 60

∴ the median lies from 20.5 to 25.5, ∴ xi= 20.5, F = 59 and f = 19

c = the range of the class interval in which median lie = 25.5- 20.5 = 5

∴ the required median, Me = 20.5 + \(\frac{60-59}{1.9}\) = 20.5 + \(\frac{5}{19}\) = 20.5 + 0.263 = 20.76

Now, rearranging the above frequency distribution table by equalising the class ranges we get,

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 12-2

We know that Mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=10 \cdot 5 ; f_0=52 ; f_{-1}=7 ; f_1=34 \text { and } c=10\)

∴ the required mode = 10.5 + \(\frac{52-7}{2 \times 52-7-34}\) x 10

= 10.5 + \(\frac{45}{104-41}\) x 10 = 10.5 + \(\frac{450}{63}\) = 10.5 + 7.143 = 17.643

Example 13. The monthly incomes (in rupees) of 70 persons are given in the following frequency distribution table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 13

Find the mode of the monthly income.

Solution: Let us draw a frequency distribution table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 13-1

According to the formula, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=299 \cdot 5, f_0=21, f_{-1}=15, f_1=7 \text { and } \mathrm{c}=100\)

∴ the required mode, \(M_0=299 \cdot 5+\frac{21-15}{2 \times 21-15-7} \times 100\)

= \(299 \cdot 5+\frac{6}{42-22} \times 100=299 \cdot 5+\frac{6}{20} \times 100=299 \cdot 5+30=329 \cdot 50\)

∴ the required Mode = Rs. 329.50

Example 14. The frequency distribution table of the prices of some houses (in Rs. 1000) is given in the following. In the table, the number of houses costing from Rs. 30,000 to Rs. 40,000. If the median of the prices of houses be Rs. 40073, then find the total number of houses.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 14

Solution: Let us draw a less-than-type cumulative frequency distribution table as follows

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 14-1

The units when changed in Rs. 1000, then the median is 40,073 and it belongs to the class intervals 40-50.

Now, the median, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_m} \times c\)

Here, \(\mathrm{Me}=40 \cdot 073, x_l=40 ; \frac{\mathrm{N}}{2}=\frac{497+f}{2} ; \mathrm{F}=196+f ; c=10 \text { and } f_m=137\)

Putting these values in the formula,

\(40 \cdot 073=40+\frac{\frac{497+f}{2}-(196+f)}{137} \times 10\)

 

or, \(\frac{40 \cdot 073-40}{10} \times 137=\frac{497}{2}+\frac{f}{2}-196-f\)

or, 1.0001 = 248.5- 196 – \(\frac{f}{2}\)

or, 1.0001 = 52.5-\(\frac{f}{2}\) or, \(\frac{f}{2}\) = 52.5 – 1.001 or, \(\frac{f}{2}\) = 51.499 or, f = 102.9998 = 103

∴ Total number of houses = N = 497 + 103 = 600

Example 15. The heights (in cm.) of 180 aged male Indians are given in the following frequency distribution table. Two of its frequencies of two class intervals are not given. The median of the heights is 163.133 Find the unknown two frequencies.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 15

Solution: Let us draw a less-than-type cumulative frequency table from the given table

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 15-1

Here, Median Me = 163.133. It lies in the class interval 161.5- 16.7.5

We know, total frequency, N = 180 = 98 + f1 + f2

∴ f1 + f2 = 180- 98 or, f1 + f2 = 82 ……(1)

Again, by the formula, \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_m} \times c\)

Here, Me = 163.133, x1 = 161.5, \(\frac{N}{2}\) = \(\frac{180}{2}\) = 90, F = 28 + f1, and c = 6 (cm.), fm = 60

Substituting these values in the formula we get,

163.133 = 161.5 + \(\frac{90-28-f_1}{60}\) x 6

or, 163-133 = 161.5 + \(\frac{62-f_1}{10}\)

or, 163.133-161.5 = \(\frac{62-f_1}{10}\)

or, 1.633 x 10 = 62- f1 or, f1 = 62- 16.33 = 45.67 = 46

Now, putting the value of f1 in (1) we get, 46 + f2 = 82 or, f2 = 82- 46 = 36

∴ f1 = 46 and f2 = 36

Example 16. The frequency distribution table of the lengths (in cm.) of the vertex of which is given below. The frequency of the class-interval 9.7- 10.4 in this table is unknown. If the mode of the lengths be 9.917 cm, then find the unknown frequency.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 16

Solution: It is seen that the mode lies in the interval 9.7- 10.4

By the formula, Mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, M0 = 9.917; x,= 9.65; f_x = 33; / = 30 and c = 0.8.

9.917 = 9.65 + \(\frac{f-33}{2 f-33-30}\) x 0.8

or, \(\frac{9.917-9.65}{0.8}=\frac{f-33}{2 f-63} or, \frac{0 \cdot 267}{0.8}=\frac{f-33}{2 f-63}\)

or, \(0.334=\frac{f-33}{2 f-63} or, 0.668 f-21 \cdot 042=f-33\)

or, \((1-0.668) \cdot f=33-21.042 or, 0.332 f=11.958\)

or, \(f=\frac{11 \cdot 958}{0 \cdot 332}=36 \cdot 01 \approx 36\).

Hence, the unknown frequency = 36.

Example 17. The frequency distribution of the heights (in inches) of 200 persons is given below. If the mode of the heights be 61.167 inches, find the unknown frequencies x and y.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 17

Solution: Here, the mid-points of the class intervals are given.

The class boundaries are 57.5- 58.5, 58.5- 59.5, 59.5-60.5,……etc.

Mode, M0 = 61.167 inches, it is in the class interval 60.5- 61.5.

Now, by the formula of Mode, \(\dot{\mathrm{M}}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(\mathrm{M}_0=61: 167, x_l=60 \cdot 5, f_0=44, f_{-1}=x, f_1=y \text { and } c=1\)

Substituting these values in the formula, we get,

\(61 \cdot 167=60 \cdot 5+\frac{44-x}{88-x-y} \times 1\)

 

or, \(61 \cdot 167-60 \cdot 5=\frac{44-x}{88-x-y}\)

or, 0.667 = \(\frac{44-x}{88-x-y}\)

or, 58.696 – 0.667x – 0.667y = 44 – x or, 58.696 – 44 = 0.667y -0.333x

or, 14.696 = 0.667 y- 0.333x….. (1)

Sum of the frequencies = 130 + x+y

∴ 200= 130 + x + y or,70 = x + y……(2)

Solving (1) and (2) we get, x = 32 and y = 38

∴ x = 32 and y = 38

The unknown frequencies x and y 32 and 38

Example 18. Find the mode and median of the following frequency distribution table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 18

Solution: Let Us draw a cumulative frequency distribution table of the above-given table

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 18-1

By the formula of median \(\mathrm{M} e=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{100}{2}\) = 50

∴ the median lies in between 60 and 65.

Here, x1 = 60, F = 44 and f0 = 30 and c = 65- 60 = 5

∴ the required median, \(\mathrm{M} e=60+\frac{50-44}{30} \times 5=60+\frac{6 \times 5}{30}\) = 60 + 1 = 61

Also, by the formula of mode, we have,

\(\mathrm{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

 

Here, \(x_l=60, f_0=30, f_{-1}=20, f_1=15\) and c = 5

Putting these values in the above formula, we get,

∴ the required mode, \(\mathrm{M}_0=60+\frac{30-20}{2 \times 30-20-15} \times 5\)

= \(60+\frac{10 \times 5}{60-35}=60+\frac{50}{25}\) = 60+ 2 = 62.

∴ the required mode = 62 cm and the required median = 61 cm.

Example 19. The frequency distribution of the marks obtained by 50 students in Math is given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 19

Find the median of the marks approximated to two decimal points.

Solution: Let us draw a cumulative frequency distribution table (less than type) of the above-given data.

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 19-1

By the formula of median \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{50}{2}\) = 25

∴ the median lies in the class-boundary 35.5- 40.5 .

∴ x1 = 35.5, F = 20 and f0 = 15

c = Range of the class boundary consisting the median = 5

Putting these values in the above formula we get, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

∴ the required median = 37.17 marks.

Example 20. The frequency distribution of the monthly incomes of 300 workers of an industry is given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 20

Determine

  1. Arithmetic Mean,
  2. Median and
  3. Mode of the above frequency distribution

Solution: 1. The mid-point of the class interval 140-150 is 145.

Let us take 145 as the origin (A) and class length (c) = 10 to be assumed as the unit.

Then, \(u=\frac{x-145}{10} \text { or, } \bar{x}=145+10 \bar{u}\)

Let us now draw the following table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 20-1

∴ \(\bar{u}=\frac{\sum u f}{\sum f}=\frac{-255}{300}=-0.85\)

\(\bar{x}\) = 145 + 10 x -0.85 = 145 -8.50= 136.50

 

or, Arithmetic mean = Rs. 136.50

2. By the formula, \(\mathrm{Me}=x_l+\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{f_0} \times c\)

Here, \(\frac{N}{2}\) = \(\frac{300}{2}\) = 150

∴ the median lies in between 130 and 140, x1= 1 30, F = 99, f0 =100 and c = 10

∴ \(\mathrm{Me}=130+\frac{150-99}{100} \times 10=130+\frac{51 \times 10}{100}\) = 130 + 5.1 = 135.10

∴ the required median = Rs. 135.10

3. Since, the highest frequency = 100

∴ the mode lies in the class-interv al (130 – 140).

By the formula of Mode, \(M_0=x_1+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_1=130, f_0=100, f_{-1}=59, f_1=41 \text { and } c=10\)

∴ Mode = \(130+\frac{100-59}{200-59-41} \times 10=130+\frac{410}{100}=130+4 \cdot 10=134 \cdot 10\)

∴ the required mode = Rs. 134.10

Example 21. Find the mean, median, and mode of the following frequency distribution

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 21

Solution: Let us draw the following table

Let us take, A = 115 as the origin, which is the mid-value of the interval (110-120) and class length, c = 10

∴ u = \(\frac{x-115}{10} \Rightarrow \bar{x}=115+10 \bar{u}\)

∴ \(\bar{u}=\frac{\sum u f}{\sum f}=\frac{231}{300}=0.77\)

∴ \(\bar{x}\) = 150 + 10 \(\bar{u}\) = 1150 + 10 x 0.77 = 115 + 7.70 = 122.70

∴ Arithmetic mean =122.70

To determine the median, let us draw the following table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 3 Median And Mode Mode Example 21-2

Here, total frequency, n = 300 and \(\frac{n}{2}\) = \(\frac{300}{2}\) = 150

The number 150 lies between 50 and 150. .*. the median lies in the interval 110- 120.

Here, xl = 110, F = 50,f0 = 100 and c = 10

∴ Median, \(\mathrm{Me}=x_l+\frac{\frac{n}{2}-\mathrm{F}}{f_0} \times c=110+\frac{150-50}{100}\) x 10= 110 + 10 = 120.

∴ The required median = 120.

Here the mode lies in the interval (110- 120), since its frequency density is the highest.

The mode, \(\mathbf{M}_0=x_l+\frac{f_0-f_{-1}}{2 f_0-f_{-1}-f_1} \times c\)

Here, \(x_l=110, f_0=100, f_{-1}=10, f_1=17 \text { and } c=10\)

∴ The mode =110 + \(\frac{100-10}{2 \times 100-10-17}\) x 10 = 110 + \(\frac{90 \times 10}{200-27}\)

= 110 + \(\frac{900}{173}=\frac{19030+900}{173}=\frac{19930}{173}\)

= 115.20 (approximated to two decimal points)

∴ The required mean = 122.70, median =120 and mode = 115.20.

 

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency

Statistics Chapter 1 Measures Of Central Tendency

In your previous classes, you have studied how different given data are summarised. Let the heights of 100 students of class X be taken. But to get a clear conception of this data, it is necessary to summarise it.

Firstly, we can prepare its frequency distribution. Secondly, to analyse the data, we can determine the different measures of the characteristics of the data.

We shall discuss here one of the characteristics of the given frequency distribution, which is a central tendency, i.e., in this chapter, we shall discuss different measures of central tendency.

You have already observed that the frequencies of any variable in the mid-positions Is higher than the frequencies of the variable in the end position.

WBBSE Solutions for Class 10 Maths

It will be seen that for the heights of the students, the number of students of middle heights is more than all. The number of very short-height or very long-height students is almost rare.

In a class, the number of very brilliant and of very dull students is always rare, most of the students are of middle memory.

In the case of incomes, it also found that the number of middle incomers is always high.

If we observe any frequency distribution keenly, we shall see that the frequencies of the first spell are low, then they gradually increase and become the highest at the mid-position and then it further gradually decreases along the end positions.

This is very characteristic of any given frequency distribution of any variable is applicable in almost all cases. Exception may arise for the mixture data.

This tendency of the frequency distributions towards the mid-positions is generally known as the central tendency.

To measure this central tendency we generally determine the average of the frequency distribution. The average is the real measure of any central tendency. You have some conceptions of different types of averages, like Arithipetic mean, Geometric mean, and Harmonic mean.

Amongst these means, we generally use the arithmetic mean as the measure of central tendency. The other two are generally used for special cases.

Except for the average, the other two which are generally used as the measures of central tendency are median and mode.

Arithmetic mean

By the term “average”, we generally mean the arithmetic mean. If the n-values of the variable x be x1,x2……….,xn, then if their arithmetic mean be mx or \(\bar{x}\) we write

\(\bar{x}=\frac{x_1+x_2+\cdots \cdots \cdots \cdots \cdots x_n}{n}=\frac{1}{n} \sum_{i=1}^n x_i \cdots \cdots \cdots(* 1)\)

 

i.e, the arithmetic mean of any variable is result of the division, when the sum of the values of the variable is divided by the number of the values.

Again, if the arithmetic mean is multiplied by the number of values we get the sum of the values of the variable, i.e.,

\(\bar{x} \times n=\sum_{i=1}^n x_i \cdots \cdots \cdots \cdots \cdots \cdots(* 2)\)

 

Again, we see that if the values of the variable be equal, then each value of the variable is equal to its arithmetic mean, i.e., \(\bar{x}\).

Thus \(\bar{x}\) is then the representative value.

Again since \(\bar{x}\) is the representative of the variable x, the frequency of x lies on \(\bar{x}\).

In this regard \(\bar{x}\) is said to be the measure of the location of x.

 

Statistics Chapter 1 Measures Of Central Tendency Examples

Example 1. Let a businessman have sold rice of 15 quintals, 19 quintals, and 20 quintals in three consecutive months. Then how many quintals of rice had he sold per month as an average?

Solution:

Given:

Let a businessman have sold rice of 15 quintals, 19 quintals, and 20 quintals in three consecutive months.

Here, let x is the variable of quantity of rice. Then the three values of x are x1, x2, and x3 the unit being quintal each.

Then the arithmetic mean of these values of x is

\(\bar{x}=\frac{x_1+x_2+x_3}{3}=\frac{15+19+20}{3} \text { quintals }=\frac{54}{3} \text { quintals }=18 \text { quintals. }\)

 

Hence, he had sold 18 quintals of rice per month.

Example 2. Let the velocity of a bicycler in 10 consecutive hours are 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 in kilometers.

Solution:

Given:

Let the velocity of a bicycler in 10 consecutive hours are 17, 25, 30, 32, 28, 24, 20, 18, 16, and 10 in kilometers

Then, the average velocity,

\(\bar{x}=\frac{x_1+x_2+\cdots \cdots+x_{10}}{10}=\frac{17+25+30+\cdots \cdots \cdot+10}{10} \mathrm{~km}=\frac{220}{10} \mathrm{~km}=22 \mathrm{~km}\)

 

Hence, his average velocity of him = 22 km/h.

Now, if the person travels 10 hours with the same velocity, he would travel a distance of 22 x 10 km = 220 km in 10 hours.

Example 3. The marks obtained by 41 students in an examination in Math are given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 3


Find the average of the marks obtained by them. Here, the total number of marks obtained

\(\sum_{i=1}^{41} x_i=15+25+32+\cdots \cdots \cdots+42=1,015\)

 

∴ the arithmetic mean, \(\bar{x}=\frac{\sum x_i}{n}=\frac{1,015}{41}\) = 24.8 (approx)

Now, for the convenient of computing of average of any variable we can transfer the origin or base of the variable x.

Thus, if we transfer the origin of x at A, then the new variable u = x- A

Here, \(\sum_{i=1}^n u_i=\sum_{i=1}^n x_i-n A \text { or, } \frac{1}{n} \sum u_i=\frac{1}{n} \sum x_i-A\)

or, \(\bar{u}=\bar{x}-A \text { or, } \bar{x}=\bar{u}+A \cdots \cdots \cdots \cdots \cdots \cdots(* 3)\)

Here, firstly the values of ui should be determined by subtracting A from the values of x, then the value of \(\bar{u}\) is to be determined.

When A is added to \(\bar{u}\),we get the average \(\bar{x}\).

Example 4. The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102, 126,85,91, 107. Find the average of amount of rupees sold per day. Here, the values of the variable x, is near the value 100 of x.

Solution:

Given:

The selling amount of money of a shopkeeper in 7 days of any week are given below (in rupees). 115,98, 102, 126,85,91, 107.

If we transfer the origin A to 100, then the values of u are 15- 2, 2, 26, – 15,- 9, 7.

∴ \(\left.\bar{u}=\frac{1}{7}(15-2+2+26-15-9+7)=\frac{50-26}{7}=\frac{24}{7}=3.43 \text { (approx. }\right)\)

∴ \(\bar{x}\) = 100 + 3.43 = 103.43 (approx)

∴the required daily selling amount = Rs 103.43

Now, let x is a discrete variable. The possible values of x1,x2……….,xn.

The frequency distribution of x are made from n values of x.

Let the frequencies of x1,x2……….,xn be f1,f2……….,fn, i.e., x1 occurs f1 times, x2 occurs f2 times,……. fn occurs n-times and f1,f2……….,fn = n.

Now, \(\sum_{i=1}^n f_i x_i=x_1 \times f_1+x_2 \times f_2+\ldots \ldots+x_n \times f_n\)

and = \(\mathrm{f}_1+\mathrm{f}_2+\ldots \ldots \ldots \ldots+\mathrm{f}_{\mathrm{n}}=\mathrm{n}\)

Arithmetic mean of \(x=\bar{x}=\frac{x_1 f_1+x_2 f_2+\cdots \cdots+x_n f_n}{f_1+f_2+\cdots \cdots+f_n}\)

= \(\frac{\sum_{i=1}^n x_i f_i}{\sum_{i=1}^n f_i}=\frac{1}{n} \sum_{i=1}^n x_i f_i \cdots \cdots \cdots \cdots \cdots \cdots(* 4)\)

This \(\bar{x}\) is the weighted mean of the variable x.

Example 5. Let a person bought 2 sharees of price Rs. 30 each, 3 sharees of price Rs. 35 each and 5 sharees of price Rs. 45 each. What is the average price of the sharees?

Solution:

Given:

Let a person bought 2 sharees of price Rs. 30 each, 3 sharees of price Rs. 35 each and 5 sharees of price Rs. 45 each.

Here, the total price of the sharees = Rs. (30 x 2 + 35 x 3 + 45 x 5)

= Rs. (60 + 105 + 225) = Rs. 390

and the total number, of sharees = 2 + 3 + 5 = 10.

∴ the average price of the sharees = Rs. \(\frac{390}{10}\) = Rs. 39.

Example 6. In a telephone exchange, the frequency distribution of the number of telephone calls per minute for any hour are given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 6

Solution: Here, at first, the following table should be prepared:

Table: Determination of the number of telephone calls per minute.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 6-2

∴ the average number of telephone calls \(\bar{x}\), where,

\(\bar{x}=\frac{\sum_{i=1}^9}{\sum_{i=1}^9 f i}=\frac{239}{60}=3 \cdot 98 \text { (approx.) }\)

Here, also the origin or base can be necessarily transferred.

Example 7. The average weight of the following frequency distribution is 117 kg.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 7

Find the value of W.

Solution: Let us at first prepare the following table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 7-1

Average weight \((\bar{x})=\frac{\sum f x}{\sum f}=\frac{920+2 \mathrm{~W}}{10}\)

Here average weight = 117 kg(Given)

∴ \(117=\frac{920+2 W}{10}\)

or, 920 + 2W = 1170 or, 2W = 1170-920

or, 2W = 250 or, W = \(\frac{250}{2}\) or, W = 125 ∴ W = 125 kg

Example 8. In a shop there are 150 balls of 5 kinds of different radius. The frequency distribution of their diameters are as follows:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 8

Find the average diameter of the balls.

Solution: Here, let 47 be the origin. Then, u = x – 47, where x = diameter in mm.

We at first prepare the following table:

Table: Determination of the average of diameters

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 8-1

∴ \(\frac{\sum u f}{\sum f}=\frac{17}{150}=0 \cdot 11 \text { (approx.) }\)

∴If the average of diameters be \((\bar{x})\), then \((\bar{x})\) = \((\bar{u})\) +47 = (0.11 + 47) mm = 47.11 mm (approx.)

In the case of continuous variable, if we want to find the average from the frequency distribution, it is quite impossible to find the correct value of the average.

Because, here we know the frequency of a certain class interval, but we do not know the individual values of each and every.

Here, we have to assume that all the values of the class interval are centered in the mid-point of that class, i.e… we should assume the frequency of the mid-point as the frequency of that class interval involved.

Thus, if 25-29 be any class interval and 10 be its frequency, then we should assume that the frequency of the mid-value 27 of the class interval 25-29 is 10.

By this assumption, we can find the approximate value of \((\bar{x})\).

Moreover, if we take the mid-value of any class interval in the middle position as the origin, the calculation becomes easier.

If the class length of the frequency distribution be equal, i.e., if the ranges of all the class intervals are equal, then it will be more convenient when the class lengths are taken as the unit of measurement.

Thus, if A be the origin and class interval c be the unit, then the new variable u will be:

u= \(\frac{x-A}{c}\) or, \(x=\mathrm{A}+c u \text { or, } x . f=A . f+c . u_f f . \text { or, } \sum x f=A \sum f+c \sum u f\)

or, \(\frac{\sum x f}{\sum f}=A+c \frac{\sum u f}{\sum f} \text { or, } \bar{x}=A+c \cdot \bar{u} \cdots \cdots \cdots \cdots \cdots \cdots(* 5)\)

Example 9. The frequency distribution of the heights (in cm) of 100 students of class X in a school. Find the average height of the students.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 9

Solution: Here, the class boundaries are 139.5- 144.5, 144.5- 149.5,…….179.5- 184.5.

The class mid-values 142, 147…….182 and class-length = 5.

The mid-value 162 of the class interval 160-164 is assumed to be the origin and 5, the class-length is assumed to be the unit, then \(u=\frac{x-162}{5} \text { and } \bar{x}=162+5 \bar{u}\)

Let us now prepare the following Table:

Table: Determination of the average heights of 100 students from frequency distribution:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 9-1

∴ \( u=\frac{\sum u \cdot f}{\sum f}=\frac{8}{100}=0.08\)

∴ \(\bar{x}=162+5 . \bar{u}\) = 162 + 5 x 0.08 = 162 + 0.40 = 162.40 cm

∴ The average height, of 100 students is 162.4 cm.

Example 10. The frequency distribution of the marks obtained by 76 students in a math exam. Find the average of marks obtained.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 10

Solution:

Table: Determination of average marks obtained in math exam of 76 students

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 10-1

∴ \(u=\frac{12}{7.6}=0 \cdot 158\)

∴ \(\bar{x}\) = 25.5 + 10 x 0. 158 = 25.5 + 1.58 = 27.08 (approx)

Properties of Arithmetic mean

1. The sum of the differences of the values of the variable from the arithmetic mean is zero.

Let the n values of x are x1,x2……….,xn.

The arithmetic mean of them is \(\bar{x}\) where \(\bar{x}=\frac{\sum x_i}{n}\)

The deviation of n-values of x from \(\bar{x}\) are \(x_1-\bar{x}, x_2-\bar{x}, \ldots \ldots \ldots x_n-\bar{x}\).

∴ \(\sum\left(x_i-\bar{x}\right)=\left(x_1-\bar{x}\right)+\left(x_2-\bar{x}\right)+\ldots \ldots+\left(x_n-\bar{x}\right)\)

= \(\left(x_1+x_2+\ldots \ldots \ldots \ldots+x_n\right)-n \bar{x}=n \bar{x}-n \bar{x}=0\)

Example 11. \(\bar{x}\) = 18 quintal.

Here, the deviations are 15- 18, 19 -18, 20- 18,

i.e.,- 3, 1, and 2. Then, the sum of- 3, 1, and 2 is zero.

Example 12.\(\bar{x}\) = 22 km.

Their deviations are – 5, 3, 8, 10, 6, 2,- 2,- 4,- 6,- 12, the sum of whose is zero.

2. If the arithmetic mean of the variable x be \(\bar{x}\) and y and x are linearly related by y = a + bx, then \(\bar{y}\) = a+bx.

Here, yi = a + bxi, i = 1, 2,…….n, then

\(\sum y_i=n a+b \sum x_i, \text { or, } \bar{y}=a+b \bar{x} \text {. }\)

 

Example 13. The arithmetic mean of the temperatures in the centigrade scale of seven different days is 32.5°C. What will be the arithmetic mean in Fahrenheit scale?

Solution:

Given:

The arithmetic mean of the temperatures in the centigrade scale of seven different days is 32.5°C.

Let the temperatures of seven days be c1,c2……….,cn in °C.

As per question, \(\overline{\mathrm{C}}=\frac{\sum \mathrm{C}_i}{7}\) = 32.5

We know that in Fahrenheit scale if the temperature be F, then \(F=32+\frac{9}{5} \bar{C}\)

∴ \(F=32+\frac{9}{5} \bar{C}\) = 32 + \(\frac{9}{2}\) x 32.5 = 32 + 9 x 6.5 = 32 + 58.5 = 90.5

∴ In the Fahrenheit scale, the average is 90-5°F.

3. If the arithmetic mean of two variables x and y be \(\bar{x}\) and \(\bar{y}\) respectively and if z =ax + by, then
\(\bar{z}\) = a\(\bar{x}\) + b \(\bar{x}\).

We have z = ax + by

∴ \(\mathrm{z}_{\mathrm{i}}=\mathrm{ax}_{\mathrm{i}}+\mathrm{by}_{\mathrm{i}}, \mathrm{i}=1,2, \ldots, \mathrm{n}, then \sum z_i=a \sum x_i+b \sum y_i\)

or, \(\frac{1}{n} \sum z_i=a \cdot \frac{1}{n} \sum x_i+b \cdot \frac{1}{n} \sum y_i\)

or, \(\bar{z}=a \bar{x}+b \bar{y}\) [because \(\frac{1}{n} \sum z_i=\bar{z}\), \(\frac{1}{n} \sum x_i=\bar{x}\) and \(\frac{1}{n} \sum y_i=\bar{y}\)

Example 14. The arithmetic means of the marks obtained by the students of a school in English and Bengali are 40.25 and 48.50 respectively. Find the averages of the sums of the marks obtained in English and Bengali.

Answer:

Given:

The arithmetic means of the marks obtained by the students of a school in English and Bengali are 40.25 and 48.50 respectively.

Here, if x be the sum of the marks obtained in English and y be that in Bengali, then their sum z will be x + y.

∴ \(\bar{z}\) = a \(\bar{x}\) + b \(\bar{y}\) = 40.25 + 48.50 = 88.75

∴ the average of the sums is 88.75.

4. For the variable x, if the arithmetic mean of two group of n1, and n2 numbers of values of \(\bar{x}_1\), and \(\bar{x}_2\) respectively, then their arithmetic mean of (n1 + n2) number of values is given by \(\bar{x}\) = \(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

Here, let the n1 number of values of x be x11, x12,…………x1n1 and let the n2 number of values of x be x21,x22,………,x2n2.

Now, if the arithmetic means of two groups be \(\bar{x}_1\) and \(\bar{x}_2\), then \(\bar{x}_1\) = \(\frac{x_{11}+x_{12}+\cdots \cdots+x_{1 n_1}}{n_1}\) and \(\bar{x}_2=\frac{x_{21}+x_{22}+\cdots \cdots+x_{2 n_2}}{n_2}\)

Also, if the collective average of them be \(\bar{x}\) then

\(\bar{x}=\frac{\left(x_{11}+x_{12}+\cdots \cdots+x_{1 n_1}\right)+\left(x_{21}+x_{22}+\cdots \cdots+\dot{x}_{2 n_2}\right)}{n_1+n_2}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

 

Example 15. There are 25 boy students and 20 girl students in the school. The average of mark in math, of 25 boy students is 52 and the average of marks in math of 20 girl-students is 48. Find the average of marks obtained by the students of the school as a whole.

Solution:

Given:

There are 25 boy students and 20 girl students in the school. The average of mark in math, of 25 boy students is 52 and the average of marks in math of 20 girl-students is 48.

Here n1= 25 and n2 = 20

\(\bar{x}_1\)= 52 and \(\bar{x}_2\) = 48.

 

∴ the collective average

\(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}=\frac{25 \times 52+20 \times 48}{25+20}\)

 

= \(\frac{1300+960}{45}=\frac{2260}{45}=50 \cdot 2 \text { (approx.) }\)

Geometric Mean

Let \(x_1, x_2, \ldots, x_n\) be n values of x.

Then if the geometric mean be \(x_g=\left(x_1 \cdot x_2 \cdot \cdots \cdots \cdots x_n\right)^{\frac{1}{n}} \cdots \cdots \cdots \cdots \cdots \cdots(* 6)\)

In order to find the average of the proportion of more than one of two variables, the geometric mean is widely used.

Let \(x_1, x_2, \ldots, x_n\) be n values of x and \(y_1, y_2, \ldots, y_n\) be n values of y.

Now, taking the proportion of x and y, let us prepare a new variable z, where z = \(\frac{x}{y}\)

The geometric mean of the variable z, is

\(z_g=\left(z_1 \cdot z_2 \ldots \ldots \ldots z_n\right)^{\frac{1}{n}}=\left(\frac{x_1}{y_1} \cdot \frac{x_2}{y_2} \ldots \ldots \frac{x_n}{y_n}\right)^{\frac{1}{n}}=\frac{\left(x_1 x_2 \ldots \ldots \ldots x_n\right)^{\frac{1}{n}}}{\left(y_1 y_2 \ldots \ldots \ldots y_n\right)^{\frac{1}{n}}}=\frac{x_g}{y_g}\)

 

To find the average of compound interests, this geometric mean is widely used

Example 16. Find the geometric mean of the following numbers: 18, 40, 25, and 45. We have a total number of numbers is 4.

Solution: ∴Geometric mean

= \((18 \times 40 \times 25 \times 45)^{\frac{1}{4}}=\{(3 \times 3 \times 2) \times(2 \times 2 \times 2 \times 5) \times(5 \times 5) \times(5 \times 3 \times 3)\}^{\frac{1}{4}}\)

= \(\left(3^4 \times 2^4 \times 5^4\right)^{\frac{1}{4}}=3 \times 2 \times 5=30\)

Harmonic mean

Let \(x_1, x_2, \ldots, x_n\) be the n-values of x are given.

If the harmonic mean of them be xn, then

\(x_n=\frac{n}{\frac{1}{x_1}+\frac{1}{x_2}+\cdots \cdots+\frac{1}{x_n}}=\frac{n}{\sum_{i=1}^n \frac{1}{x_i}} \cdots \cdots \cdots \cdots \cdots \cdots(* 7)\)

 

Sometimes, the given variable is given as the rate of some objects.

For example, velocity per hour, price per kilogram, weight per cubic foot, etc.

In these cases, the harmonic mean is used.

Example 17. A train travels the first 30 miles at a speed of 15 miles/hour and the next 30 miles at a speed of 30 miles/hour. Find the average velocity of the train.

Solution:

Given:

A train travels the first 30 miles at a speed of 15 miles/hour and the next 30 miles at a speed of 30 miles/hour.

Here, the total distance = (30 + 30) miles = 60 miles.

Total times required = \(\left(\frac{30}{15}+\frac{30}{30}\right)\) hours =(2 + 1) hours = 3 hours.

∴ the average velocity of the train = \(\frac{60}{3}\) miles / hour = 20 miles / hour.

By using the formula of harmonic mean, we can also find the required average.

The train has travelled same distance with two types of velocities.

These two velocities are 15 miles/hour and 30. miles/hour. The harmonic mean of these two types of Velocities

= \(\frac{2}{\frac{1}{15}+\frac{1}{30}}=\frac{2}{\frac{2+1}{30}}=2 \times \frac{30}{3}=20\)

∴ the required average = 20 miles/hour

Example 18. A shopkeeper bought rice of Rs. 30 at a rate of Rs 1.5 per kg and of Rs 30 at a rate of Rs. 4 per kg. Then what is the value of mixed rice per kg?

Solution:

Given:

A shopkeeper bought rice of Rs. 30 at a rate of Rs 1.5 per kg and of Rs 30 at a rate of Rs. 4 per kg.

Here, the total amount of buying price of the shopkeeper = Rs. (30 + 30) = Rs. 60.

The quantity of the total rice bought

= \(\frac{30}{1 \cdot 5}+\frac{30}{4} \mathrm{~kg}=20+7 \frac{1}{2} \mathrm{~kg}=27 \frac{1}{2} \mathrm{~kg}\)

∴ the price of the mixed rice per kg = Rs \(\frac{60}{27 \frac{1}{2}}\)

= Rs. \(\frac{60}{\frac{55}{2}}\) = Rs \(\frac{24}{11}\) = Rs 2.18 (approx.)

∴ The required rate of price = Rs. 2.18 per kg.

We can also find the result by using the harmonic mean.

According to the formula, the H.M.

= \(\frac{2}{\frac{1}{1 \cdot 5}+\frac{1}{4}}=\frac{2}{\frac{1}{1 \frac{1}{2}}+\frac{1}{4}}=\frac{2}{\frac{1}{\frac{3}{2}}+\frac{1}{4}}=\frac{2}{\frac{2}{3}+\frac{1}{4}}\)

= \(\frac{2}{\frac{8+3}{12}}=\frac{2 \times 12}{11}=\frac{24}{11}=2 \cdot 18\)

∴ the average price of rice = Rs. 2.18 per. kg

Example 19. The frequency distribution of the monthly salaries of 80 workers of a company are given below. Find the arithmetic mean of the salaries of the workers.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 19

Solution: Here, we have to prepare class boundaries and class mid-values.

Due to the different lengths of the class intervals, by changing the origin and order, it is impossible to find out the average.

So, we have to prepare the following table at first:

Determination of average of the salaries of 80 workers from the given frequency distribution

∴ \(\bar{x}=\frac{\sum x f}{\sum f}=\frac{32390}{80}=404 \cdot 87\)

∴ the average of monthly salary of 80 workers = Rs. 404.87

Example 20. There are two groups A and B in a shop. In group A, 50 toys have been sold the average price of each toy being Rs. 15. In group B there also have been sold 30 toys, the average price of which is Rs. 20 each. Find the average of the average selling price of the total 80 toys sold in the shop.

Solution:

Given:

There are two groups A and B in a shop. In group A, 50 toys have been sold the average price of each toy being Rs. 15. In group B there also have been sold 30 toys, the average price of which is Rs. 20 each.

Here, the total selling price of 50 toys in group A = Rs. 50 x 15 = Rs. 750

The total selling price of 30 toys in group B = Rs. 30 x 20 = Rs. 600

∴ The total selling price = Rs. (750 + 600) = Rs. 1350

∴ the required average = Rs. \(\frac{1350}{50+30}\) = Rs.\(\frac{1350}{80}\) = Rs. 16.87 (approx.)

Example 21. The frequency distribution of the degree of the drugs of 110 patients to be cured are given below Here the number of patients being cured by taking 20 tablets and the number of patients being cured by taking 32 tablets are unknown. But it is known that the number of tablets required to be cured per patient is 20 tablets. Find the two unknown frequencies.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 21

Solution: Let the unknown frequencies be f1 and f2.

Also, let the origin of the variable (here tablets) be 20 and the unit of degree be 4, the class length, then

\(u=\frac{x-20}{4} \text { and } x=20+4 \bar{u}\)

 

To find the average, we prepare the following table

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 21-1

The total frequency = 110, ∴ 110 = 90 + f1 + f2 or, f1+ f2 = 20…….(1)

Also, the number of average tablets = 20

∴ 20= 20 + 4 . \(\bar{u}\) or, 0 = 4. \(\frac{\left(-42+3 f_2\right)}{110}\)

or,- 42 + 3f2 = 0 or, 3f2 = 42 or, f2 = \(\frac{42}{3}\) = 14

∴ from (1), we get, f1 + 14 = 20 or, = 20- 14 = 6

∴ the unknown two frequencies are 6 and 14.

Example 22. The frequency distributions of the weekly wages of some laborers of industry are given below: The number of laborers having Rs 25 weekly wages is unknown. But it is known that the average weekly wages of a labourer is Rs 27.50. Find the unknown frequency.

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 22

Solution: Here, assuming 30 as the origin and the class length 5 as the unit of degree, we get, the new variable u,

where, \(u=\frac{x-30}{5}\), i.e., x=30+5 \(\bar{u}\)

Let the unknown frequency be f. Then we prepare the following table

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 22-1

∴ \(\bar{u}=\frac{\sum u \times f}{\sum f}=\frac{-17-f}{52+f} \text { and } \bar{x}=30+5 \cdot \frac{-17-f}{52+f}\)

or, 27.5 = 30.5 . \(\frac{-17-f}{52+f}\) [x = Rs. 27.5]

or,-0.5 = \(\frac{-17-f}{52+f}\) or, 0.5(52 + f) = 17 + f, or, 26 + 0.5 f = 17 + f.

or, 26-17 = f – 0.5f, or, 9 = 0.5f or, f = 18, which is the required frequency.

Example 23. In a business house, the weekly salaries of 30 employees are given below:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 23

The rate of bonus of that house are given in the following table :

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 23-1

Find the arithmetic mean of. the bonus of the employees.

Solution: We have to prepare, firstly the frequency distribution table.

In this table, the class intervals are 61-75, 76-90,91-105, etc.

The frequencies of these intervals are f1, f2, f3,……..etc. and the rate of bonus etc. are y1, y2, y3, … etc.

Then the arithmetic mean of bonus, \(\bar{y}=\frac{\sum y_i f_i}{\sum f}\)

To find the arithmetic mean, let us prepare the frequency table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 23-2

In the case of the variable, let us assume 1 00 as the origin and class length 254 as the unit then \(u=\frac{y-100}{25}\) and \(\bar{y}=100+25 \cdot \bar{u}\).

Here, \(\bar{u}=\frac{\sum u \cdot f}{\sum f}=\frac{24}{30}=\frac{4}{5}\)

∴ \(\bar{y}\) = 25. \(\frac{4}{5}\) = Rs. (100 +20) = Rs. 120

∴ the average of bonus is Rs. 120

Example 24. There are 50 workers in a small industry. The daily income of them is Rs. 9. Amongst them, the daily income of the 30 workers, working in the morning is Rs. 8. Then find the daily average income of the workers working in the evening.

Solution:

Given:

There are 50 workers in a small industry. The daily income of them is Rs. 9. Amongst them, the daily income of the 30 workers, working in the morning is Rs. 8.

Here, the total number of workers, n = 50

The number of workers in the morning n1 = 30

∴ The numbers of workers in the evening n2 = n- n1 = 50- 30 = 20

The average daily of the total workers = Rs. 9

The average daily income of the workers in the morning, \(\bar{x}_1\) = Rs. 8

According to the law of collective average,

\(\bar{x}=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2}\)

 

∴ \(9=\frac{30 \times 8+20 \bar{x}_2}{30+20} \text { or, } 9=\frac{240+20 \bar{x}_2}{50}\)

or, 450 = 240 + 20 \(\bar{x}_2\) or, 20 \(\bar{x}_2\) = 450-240

or, 20 \(\bar{x}_2\) =210 or, 20 \(\bar{x}_2\) = 210 or, \(\bar{x}_2\) = \(\frac{210}{20}\) = 10.50

∴ The daily income of the workers in the evening = Rs. 10.50.

Example 25. Find the day of the average absence of each and every student from the following given table:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 25

Solution: Here, the upper boundaries of the class intervals and less-than-type cumulative frequencies are given.

First, the following table is to be prepared:

WBBSE Solutions For Class 10 Maths Statistics Chapter 1 Measures Of Central Tendency Example 25-1

∴ \(\bar{u}=\frac{\sum u . f}{\sum f}=\frac{-40}{630}=-\frac{4}{63}=-0.0635 \text { (approx.) }\)

∴ \(\bar{x}\) = 12 + 5 \(\bar{u}\) = 12 + 5(-0.0635 = 12-0.3175 = 11.6825

∴ The average absence of the students = 11.68 days.

Example 26. The arithmetic mean of 200 values of a variable is 50. Later on, it is found that two values 92 and 8 have been taken wrongly instead of 192 and 88 respectively. After correction, find the actual arithmetic mean of the given values.

Solution:

Given:

The arithmetic mean of 200 values of a variable is 50. Later on, it is found that two values 92 and 8 have been taken wrongly instead of 192 and 88 respectively.

Here, the summation of the values of the variable = \(=\sum_{i=1}^{200} x_i=200 \bar{x}\) = 200 x 50 = 10000

Taking correct values, the summation = 10000-92-8+ 192 + 88= 10180

∴  the corrected arithmetic mean = \(\frac{10180}{200}\) = 50.9

Example 27. The average of the weights of 150 students is 50 kilogram. The average of the weights of the boy students is 55 kilograms and the average of the weights of the girl students is 42.5 kilogram. Find the number of boy and girl students.

Solution:

Given:

The average of the weights of 150 students is 50 kilogram. The average of the weights of the boy students is 55 kilograms and the average of the weights of the girl students is 42.5 kilogram.

Here, the average of the weights of the total students is \(\bar{x}\) = 50 kilogram

Total number of students n = 150

The average of weights of the boy students, \(\bar{x}_1\) = 55 kilogram

The average of weights of the girl students, \(\bar{x}_2\)=42.5 kilogram

We have to find the number of boy students, and the number of girl students.

It is known that n = n1 + n2 or, 150 = n1 + n2……..(1)

Putting these values in the formula of collective arithmetic mean, we get

\(50=\frac{n_1 \bar{x}_1+n_2 \bar{x}_2}{n_1+n_2} \text { or, } 50=\frac{n_1 \times 55+n_2 \times 42.5}{150}\)

 

or, 7500 = 55n1 + 42.5n2………(2)

Solving (1) and (2), we get n1 = 90 and n2 = 60

∴ The number of boy students = 90 and the number.of girl students = 60

Example 28. If a and b be two positive numbers, then prove that A.M ≥ G.M. ≥ H.M. where

A.M. = Arithmetic mean
G.M. = Geometric mean
H.M = Harmonic mean

Solution: If a and b be two positive numbers, then

A.M. = \(\frac{a+b}{2}, \mathrm{G} \cdot \mathrm{M} .=(a b)^{\frac{1}{2}}=\sqrt{a b}\) and

H.M = \(\frac{2}{\frac{1}{a}+\frac{1}{b}}=\frac{2}{\frac{a+b}{a b}}=\frac{2 a b}{a+b}\)

Now, A.M.-G.M.= \(\frac{a+b}{2}-\sqrt{a b}=\frac{1}{2}(a+b-2 \sqrt{a b})=\frac{1}{2}(\sqrt{a}-\sqrt{b})^2\)

We know (√a- √b) may be positive or negative. But its square is always positive.

The value of (√a- √b) is 0 when √a = √b.

∴ \(\frac{1}{2}(\sqrt{a}-\sqrt{b})^2 \geq 0\)

or, A.M.- G.M ≥ 0. or. A.M > G.M……(1)

Substituting \(\frac{1}{a}\) and \(\frac{1}{a}\) instead of a and b respectively and the A.M. > G.M

∴ \(\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right) \geq\left(\frac{1}{a} \cdot \frac{1}{b}\right)^{\frac{1}{2}}\)

or, \(\frac{a+b}{2 a b} \geq \frac{1}{\sqrt{a b}}\),

or, \(\sqrt{a b} \geq \frac{2 a b}{a+b}\)

∴ G.M. ≥ H.M,………(2)

From the above relations (1) and (2). we get. A.M. ≥ G.M. ≥ H.M.

∴ A.M.≥ G.M ≥ H.M (Proved)

Example 29. The AM of two positive numbers is 25 and their GM is 15. Find the two numbers.

Solution:

Given:

The AM of two positive numbers is 25 and their GM is 15.

Let the two positive numbers be a and b

∴ their AM = \(\mathrm{AM}=\frac{a+b}{2}\) and GM = \((a b)^{\frac{1}{2}}=\sqrt{a b}\)

As per question, \(\mathrm{AM}=\frac{a+b}{2}\) or, a + b = 50………(1)

and =15 or, ab = 225……..(2)

From (1) we get, a = 50- b

∴ from (2) we get (50- b) b = 225, or, 50b- b2 = 225

or, b2– 50b+ 225 = 0 or, b2– 5b- 45b + 225 = 0

or, b (b- 5)- 45 (b- 5) = 0 or, (b- 5) (b- 45) = 0

∴ either b-5 = 0or, b = 5 or, b- 45 = 0 or, b = 45

∴ b = 5 and b = 45

∴ the required numbers are 5 and 45.

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances

What is line of sight?

The straight line drawn from the eye of an observer in the object which is viewed by the observer is called line of sight.

In beside the eye of an observer is at the point O and A is the position of the object viewed by the observer.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Line Of Sight

 

By joining the points O and A, we get a straight line OA.

This straight line OA is called the line of sight.

Since the eye-sight always passes along a straight line, the line of sight can never be a curved line.

WBBSE Solutions for Class 10 Maths

What is angle of elevation?

The angle which is made by the line of sight with the horizontal line is called the angle of elevation.

In an angle of elevation an observer always looks from downwards to the upwards.

The angle of elevation may take any values between 0° and 360°.

The angle θ is called the angle of elevation, since it is formed by the line of sight with the horizontal line.

Class 10 Maths Solutions Wbbse

What is angle of depression?

If an observer is looking down to an object, then the angle produced by his line of sight with the horizontal line or with the line parallel to the horizontal line is called the angle of depression.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Angle Of Depression

 

The eye of an observer is at O and the position of the object is at A.

OC is a straight line parallel to the horizontal line BA. Then OA is the line of sight of the observer which makes an angle θ with the line OC.

Hence θ is the angle of depression.

In angle of depression an observer looks from upwards to downwards.

Obviously, ∠OAB = θ, since OC || BA and OA is a transversal of OC and BA,

∴ ∠AOC = ∠OAB [alternate angles]

∴ ∠OAB = θ.

Application of trigonometric ratios in practical problems:

In the following examples we have discussed different types of applications of trigonometric ratios

Class 10 Maths Solutions Wbbse

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Questions

Example 1. If the angle of elevation of the top of a tower from a distance of 10 metres from its foot is 60°, then the height of the tower is

  1. 5√3 metres
  2. 10√3 metres
  3. \(\frac{10}{\sqrt{3}}\) metres
  4. None of these

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Question Example 1

 

Solution: Let AB is the tower and C is at a distance of 10 metres from its foot B.

As per the question, BC = 10 metres and ∠ACB = 60°.

Now, from the right-angled triangle ABC we get,

tan 60° = \(\frac{AB}{BC}\) [by definition]

or, √3 = \(\frac{AB}{10}\) or, AB = 10√3 metres.

∴ the height of the tower is 10√3 metres.

Hence 2. 10√3 metres is correct.

Example 2. In the adjoining, the value of θ is

  1. 25°
  2. 30°
  3. 35°
  4. 45°

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Question Example 2

 

Solution: The given figure is a right-angled triangle of which the length of the perpendicular is 5 metres and the length of the base is 5√3 metres.

∴ \(\tan \theta=\frac{5 \text { metres }}{5 \sqrt{3} \text { metres }}\) [by definition]

or, \(\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}\)

⇒ θ = 30°

Hence 2. 30° is correct.

Class 10 Maths Solutions Wbbse

Example 3. At what angle an observer observes a box lying on ground from the roof of three-storied building, so that the height of building is equal to the distance of the box from the building?

  1. 25°
  2. 30°
  3. 45°
  4. 55°

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Question Example 3

 

Solution: Let the box be at C on the ground and the observer observes it from point A of the roof of building AB.

As per the question, AB = BC, AE || BC, ∴ ∠CAE = ∠ACB = θ (let).

∴ from the right-angled ΔABC we get,

tan θ = \(\frac{AB}{BC}\) or, tan θ = \(\frac{AB}{AB}\)

or, tan θ = 1 or, tanθ = tan 45° ⇒ θ = 45°,

Hence 3. 45° is correct.

Class 10 Maths Solutions Wbbse

Example 4. Height of a coconut tree is 100√3 metres. The angle of elevation of the top of the coconut tree from a point at a distance of 100 metres of foot of the coconut tree is

  1. 40°
  2. 45°
  3. 55°
  4. 60°

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Question Example 4

 

Solution: Let the angle of elevation is θ and AB is the coconut tree.

As per question, AB = 100√3 metres and BC = 100 metres

∴ from the right-angled ΔABC, we get,

tan θ = \(\frac{AB}{BC}\)

or, tan θ = \(\frac{100 \sqrt{3}}{100}\) or, tan θ = √3

or, tan θ = tan 60°

⇒ θ = 60°

Hence 4. 60° is correct.

Class 10 Maths Solutions Wbbse

Example 5. If the length of the shadow on the ground of a palm tree is √3 times of its height, the angle of elevation of the sun is

  1. 30°
  2. 45°
  3. 55°
  4. 65°

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Multiple Choice Question Example 5

 

Solution: Let the angle of elevation of the sun is 0 and AB is the palm tree, the shadow of which is BC.

As per question, BC = √3 AB

Now, from the right-angled ΔABC We get,

tan θ= \(\frac{AB}{BC}\)

or, tan θ = \(\frac{\mathrm{AB}}{\sqrt{3} \mathrm{AB}}\) or, tan θ = \(\frac{1}{\sqrt{3}}\)

or, tan θ = tan 30° ⇒ θ = 30°

Hence 1. 30° is correct.

 

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances True Or False

Example 1. In ΔPQR, ∠Q = 90°, If PQ = QR, then ∠R = 60°.

Solution: False

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances True Or False Example 1

 

Since, let ∠R = θ, then tan θ = \(\frac{PQ}{QR}\)

= \(\frac{PQ}{QR}\) [∵ PQ = QR]

⇒ tan θ = 1

⇒ tan θ = tan 45°

⇒ θ = 45°

Example 2. AB is the height of a tower, BC is the base, the angle of depression from a point A at the point C is ∠DAC; So, ∠DAC = ∠ACB.

Solution: True

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances True Or False Example 2

 

Since AD || BC and AC is their transversal,

∴ ∠DAC and ∠ACB are alternate angles.

∴ ∠DAC = ∠ACB.

 

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Fill In The Blanks

Example 1. If the sun’s angle of elevation increases from 35° to 70°, the length of shadow of a tower _______

Solution: decreases

Example 2. If the angle of elevation of sun is 40°, the length of shadow and length of tower are _______

Solution: equal

Example 3. If the angle of elevation of sun is _____ than 50°, the length of shadow of post will be less than the height of post.

Solution: greater

 

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Type Questions

Example 1. If the angle of elevation of a kite is 60° and the length of thread is 20>√3 metres, calculate the height of kite above the ground.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 1

 

Let the height of the kite be AB = h metre above the ground and let AC be the thread.

As per question, AC = 20√3 metres and ∠ACB = angle of elevation of the kite at A = 60°.

Now, from the right-angled triangle ABC,

sin ∠ACB = sin 60° = \(\frac{AB}{AC}\) [by definition of sinθ]

or, \(\frac{\sqrt{3}}{2}=\frac{h}{20 \sqrt{3}}\)

or, 2h = 20 x 3

or, \(\quad h=\frac{20 \times 3}{2}=30\)

Hence the required height of the kite above the ground = 30 metres.

Example 2. AC is the hypotenuse with length of 80 metres of a right-angled triangle ABC and if AB = 40√3 metres, then find the value of ∠C.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 2

 

Let ∠C = 0.

As per the question, AB = 40√3 m and AC = 80 m.

Now, from the right-angled triangle ABC, sin θ = \(\frac{AB}{AC}\) [by definition]

⇒ \(\sin \theta=\frac{40 \sqrt{3}}{80}\)

⇒ \(\sin \theta=\frac{\sqrt{3}}{2}\)

⇒ sin θ = sin 60°  ⇒ θ = 60°

Hence the value of ∠C = 60°.

Example 3. A tower breaks due to storm and its top touches the ground in such a manner that the distance from the top of the tower to the base of the tower and present height are equal. Let us calculate how much angle is made by the top of the tower with the base.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 3

 

Let AB be the tower and it breaks at the point C such that BD = BC, where the top of the tower meets the ground at the point D.

Let ∠BDC = θ.

Now, from the right-angled ΔBCD we get,

tan θ = \(\frac{BC}{BD}\) [by definition]

⇒ tan θ = \(\frac{BC}{BC}\) [BD = BC (given)]

⇒ tan θ = 1 = tan 45° ⇒ θ = 45°

Hence the top of the tree makes an angle of 45° with the base.

Example 4. In the right-angled triangle PQR, ∠Q = 90°, S is such a point on PQ that PQ: QR: QS = √3 : 1: 1, then find the value of ∠PRS.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 4

 

Let ∠QRS = θ.

Given that PQ: QR: QS = √3: 1: 1,

i.e., QR : QS = 1 : 1

⇒ \(\frac{\mathrm{QR}}{\mathrm{QS}}=1 \Rightarrow \frac{\mathrm{QS}}{\mathrm{QR}}=1\)…….(1)

Now, in right-angled triangle QRS, we get,

tan θ = \(\frac{QS}{QR}\) [by definition]

⇒ tan θ = 1 [ from (1) ]

⇒ tan θ = tan 45°

⇒ θ = 45°

Again, given that PQ: QR = √3: 1

or, \(\frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\sqrt{3}}{1}=\sqrt{3}\)

Now, from the right-angled triangle PQR, we get,

tan ∠PRQ = \(\frac{\mathrm{PQ}}{\mathrm{QR}}\)

⇒ tan ∠PRQ = √3= tan 60°

⇒ ∠PRQ = 60°.

∴ ∠PRS = PRQ – ∠QRS

= 60° – 45° = 15°

Hence the required value of ∠PRS = 15°.

Example 5. If the ratio between length of shadow of a tree and height of tree is √3: 1, then find the angle of elevation of the sun.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 5

 

Let AB be the tree and BC be its shadow, when the angle of elevation of the sun is θ.

As per the question,. BC : AB = √3 : 1

⇒ \(\frac{\mathrm{BC}}{\mathrm{AB}}=\sqrt{3} \Rightarrow \frac{\mathrm{AB}}{\mathrm{BC}}=\frac{1}{\sqrt{3}}\)……….(1)

Now, from the right-angled triangle ABC,

tan θ = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

⇒ tan θ = \(\frac{1}{\sqrt{3}}\) [ from (1) ]

⇒ tan θ = tan 30°

⇒ θ= 30°

Hence the angle of elevatio’n of the sun is 30°.

Example 6. The length of the shadow of a tree is 9 metres when the sun’s angle of elevation is 30°. Calculate the height of the tree.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 6

 

Let AB be the tree and BC be its shadow.

As per question, BC = 9 m and ∠ACB = 30°.

Now, from the right-angled triangle ABC, we get,

tan 30° = \(\frac{AB}{BC}\) [by defination]

or, \(\frac{1}{\sqrt{3}}=\frac{A B}{9}\)

or, √3AB = 9

or, AB = \(\frac{9}{\sqrt{3}}\) = 3√3

Hence the height of the tree = 3√3 metres.

Example 7. A tower stands on the bank of a river. A post is fixed in the earth on the other bank just opposite to the tower. On moving 7√3 metres from the post along the bank, it is found that the tower makes an angle of 60° at that point with respect to this bank. Find the width of the river.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Short Answer Question Example 7

 

Let the width of the river be AB. A is the position of the tower and B is the position of the post.

As per the question,

if BC = 7√3 m, then ∠ACB = 60°.

Now, from the right-angled triangle ABC, we get,

tan60°= \(\frac{AB}{BC}\)

or, \(\sqrt{3}=\frac{A B}{7 \sqrt{3}}\)

or, AB = √3 x 7 √3 or, AB = 21

Hence the width of the river is 21 metres.

 

Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Type Questions

Example 1. Two towers stand just on the opposite sides of a road. A ladder that stands against the pillar of the first towyer is at a distance of 6 metres from the tower and makes an angle of 30° with the horizontal line. But if the ladder stands against the pillar of the second tower keeping its foot at the same point, it makes an angle of 60° with the horizontal line.

  1. Calculate the distance of the foot of the ladder from the foot of the pillar of the second tower.
  2. Find the width of the road.
  3. Find the height where the top of the ladder is fixed against the pillar of the second tower.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 1

 

Let AB and CD be two towers so that BD is the width C of the road.

The ladder PQ makes an angle of 30° when it is against the first tower AB and the same ladder PR makes an angle of 60° when it is R against the second tower CD.

∴ ∠BPQ = 30° and ∠DPR = 60°

As per question, BP = 6 metres

Now, from the right-angled triangle BPQ we get,

cos 30° = \(\frac{BP}{PQ}\) [by definition]

or, \(\frac{\sqrt{3}}{2}=\frac{6}{P Q} \quad \text { or, } \sqrt{3} \mathrm{PQ}=12 \quad \text { or, } \quad P Q=\frac{12}{\sqrt{3}}=\frac{4 \sqrt{3} \cdot \sqrt{3}}{\sqrt{3}}=4 \sqrt{3}\)

Hence the length of the ladder = 4√3 metres.

Again, from the right-angled triangle DPR we get, cos 60°= \(\frac{PD}{PQ}\) [by definition]

or, \(\frac{1}{2}\) = \(\frac{PD}{PQ}\) [PR = PQ]

or, \(\frac{1}{2}=\frac{P D}{4 \sqrt{3}} \quad \text { or, } \quad 2 P D=4 \sqrt{3} \quad \text { or, } \quad P D=\frac{4 \sqrt{3}}{2} . \text { or, } \quad P D=2 \sqrt{3}\)

∴ BD = BP + PD = (6 + 2√3) m = 2 (3 + √3) m

Hence the width, of the road = 2 (3 + √3) m .

Also, from the right-angled triangle PDR we get, sin 60° = \(\frac{DR}{PR}\) [ by definition ]

or, \(\frac{\sqrt{3}}{2}=\frac{\mathrm{DR}}{\mathrm{PQ}}\) [PR = PQ]

or, \(\frac{\sqrt{3}}{2}=\frac{\mathrm{DR}}{4 \sqrt{3}}\)

or, 2DR = 12 or, DR = \(\frac{12}{2}\) = 6

Hence the ladder is fixed against the pillar of the second tower at a height of 6 metres above the ground.

Hence,

  1. the distance of the foot of the ladder from the foot of the pillar of the second tower is 2√3 m.
  2. the width of the road is 2 (3 + √3) m.
  3. the height where the top of the ladder is fixed against the pillar of the second tower is 6 m.

Example 2. If the angle of elevation of the top of a house from a point on the horizontal plane passing through the foot of the house is 60° and the angle of elevation from another point on the same plane at a distance of 24 metres away from the first point is 30°. Find the height of the house. [√3 = 1.732 (approx). ]

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 2

 

Let AB be the house.

The angle of elevation of the top A of it from the point C is 60° and from the point D, 24 metres away from the point C is 30°.

∴ ∠ACB = 30° and ∠ADB = 60°,

Now from the right-angled triangle ABC, we get,

tan 60° = \(\frac{AB}{BC}\) [by definition]

or, √3 = \(\frac{AB}{BC}\)

or, AB = √3 BC………(1)

Again, from the right-angled triangle ABD, we get,

tan 30°= \(\frac{AB}{BC}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{A B}{B C+C D}\)

or, √3AB = BC + 24 [ ∵ CD = 24 m]

or, √3AB- BC = 24

or, √3AB- \(\frac{AB}{\sqrt{3}}\) = 24 [from( 1 )

or, AB \(\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)\) = 24

or, AB \(\left(\frac{3-1}{\sqrt{3}}\right)\) = 24

or, AB x \(\frac{2}{\sqrt{3}} = 24\) or, AB = \(\frac{24 \sqrt{3}}{2}\)

or, AB = 12√3 = 12 x 1.732 (approx.) = 20.784 (approx.)

∴ Height of the house = 20.784 metres (approx.)

Example 3. When the top of a tower is seen from a point on the roof of the building of 9√3 metre height, the angle of elevation is 30°. If the distance between them is 30 metres, find the height of the tower.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 3

 

Let AB be the tower and CD be the building.

CE ⊥ AB, ∴ CE = BD and CD = BE

As per question, ∠ACE = 30°

Now, from the right-angled triangle ACE we get,

tan 30° = \(\frac{AE}{CE}\)

or, \(\frac{1}{\sqrt{3}}=\frac{\mathrm{AE}}{\mathrm{CE}}\)

or, √3AE = CE or, √3AE = BD

or, √3AE = 30 or, AE = \(\frac{30}{\sqrt{3}}\) = 10√3

Now, AB = AE + BE

=10√3 + 9√3 =19√3

Hence the height of the tower = 19√3 metres.

Example 4. The length of the shadow of a coconut tree becomes 3 metres smaller when the angle of elevation of the sun increases from 45° to 60°. Calculate the height of the coconut tree. [ Let √3 = 1.732 (approx.)]

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 4

 

Let BD be the shadow of AB when the angle of elevation is 45° and BC be the length of the shadow of the coconut tree AB when the angle of elevation is 60°.

Now, from the right-angled triangle ABC we get,

tan 60° = \(\frac{AB}{BC}\) [by definition]

or, √3 = \(\frac{AB}{BC}\) or, AB = √3 BC

or, BC = \(\frac{\mathrm{AB}}{\sqrt{3}}\)………(1)

Again, from the right-angled triangle ABD, we get,

tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) [by definition]

or, 1 = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) or, AB = BD or, AB = BC + CD

or, AB = \(\frac{\mathrm{AB}}{\sqrt{3}}\) + CD or, AB – \(\frac{\mathrm{AB}}{\sqrt{3}}\) = CD

or, AB (1 – \(\frac{1}{\sqrt{3}}\)) = CD

or, AB\(\left(\frac{\sqrt{3}-1}{\sqrt{3}}\right)\) = 3 [∵ CD = 3 metres] or, AB = \(\frac{3 \times \sqrt{3}}{\sqrt{3}-1}\)

or, AB = \(\frac{3 \sqrt{3}}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\) or, AB = \(\frac{3 \sqrt{3}(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2}\)

or, AB = \(\frac{3 \sqrt{3}(\sqrt{3}+1)}{3-1}\)

or, \(\frac{3 \sqrt{3}(\sqrt{3}+1)}{2}\)

or, AB = \(\frac{9+3 \sqrt{3}}{2}=\frac{9+3 \times 1 \cdot 732}{2}\)

= \(\frac{9+5 \cdot 196}{2}=\frac{14 \cdot 196}{2}\) = 7.098

Hence the height of the coconut tree = 7.098 metres (approx.)

Example 5. The bottom of a three storied building and bottom of two lamp posts are on the same straight line and the angles of depression from three storied building at the bottom of two lamp posts are 60° and 30° respectively. If the distance of the points at the bottom of the three storied building and the bottom of the first lamp post is 200 metres, calculate how far will the 2nd lamp post from the building be and what will the height of building be.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 5

 

Let AB be the three-storied building and D and C are the bottom of the two lamp posts and BD = 200 metres.

Now, from the right-angled triangie ABD we get,

tan 60°= \(\frac{AB}{BD}\) or, √3 = \(\frac{\mathrm{AB}}{200}\) or, AB = 200√3

Also, from the right-angled triangle ABC, we get,

tan30° = \(\frac{AB}{BC}\) or, \(\frac{1}{\sqrt{3}}=\frac{A B}{B C}\)

or, BC = √3AB = √3 x 200 √3 = 600

Hence the distance of other ship from the lighthouse is 600 metres and the height of the lighthouse is 200 √3 metres.

Example 6. A pilot of a helicopter observes that Sealdah station is at. one side of the helicopter and Manument is just on the opposite side. The angles of depression of Sealdah station and Manument from the pilot of the helicopter are 60° and 30° respectively. If the helicopter is at a height of 235>/3 metres at that time, find the distance between Sealdah station and Manument.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 6

 

Let the position of Sealdah station be H and the position of Manument be S. OD is the height of the helicopter.

As per question OD = 235√3 metres,

∠EOH = ∠OHD = 60° and ∠FOS = ∠OSD = 30°

Now, from the right-angled triangle OHD, we get,

tan 60° = \(\frac{OD}{HD}\) [by definition]

\(\frac{1}{\sqrt{3}}=\frac{235 \sqrt{3}}{\mathrm{DS}}\) or, HD = 235

 

Also, from the right-angled triangle OSD, we get,

tan 30° = \(\frac{OD}{DS}\) [by defnition]

or, \(\frac{1}{\sqrt{3}}=\frac{235 \sqrt{3}}{\mathrm{DS}}\)

or, DS = 235 x √3 x √3

or, DS = 235 x 3 = 705

Now, HS = HD + DS = 235 + 705 = 940

Hence the distance between Sealdah station and Manument is 940 metres.

Example 7. From a point on the roof of a building the angle of elevation of the top of a tower and that of angle of depression of the foot of the tower are 60° and 30° respectively. If the mheight of the building is 16 metres, calculate the height of the tower and the distance of the building from the tower.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 7

 

Let AB be the tower and CD be the building.

∴ DB ⊥ AB  ∴ DE = BC and CD = BE = 16 m

As per questions, ∠ADE = 60° and ∠BDE = 30°

Now, from the right-angled triangle BDE, we get,

tan 30° = \(\frac{BE}{DE}\) [by definition]

or, \(\frac{\sqrt{3}}{2}=\frac{A B}{300}\) or, DE = 16√3…….(1)

Again, from the right-angled triangle ADE we get,

tan 60° = \(\frac{A B}{D E}\)

or, \(\sqrt{3}=\frac{A E}{16 \sqrt{3}}\) or, AE = 48

∴ AB = AE + BE = (48 + 16) m. = 64 m.

Hence the height of the tower is 64 m. and the distance of the building from the tower is 16√3 m.

Example 8. Shib is flying a kite having the length of thread 300 metres, when the thread makes an angle 60° with the horizontal line and when the thread makes an angle of 45° with the horizontal line. Calculate in each case what is the height of kite from the ground. Also find in which of the two cases will the kite be at a greater height from the other.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 8

 

Let the height of the kite when the thread of length 300 metres makes an angle 60° with the horizontal line is AB and when 45° is CD.

As per the question, OA = OC = 300 m.

Now, from the right-angled triangle OAB, we get,

sin 60° = \(\frac{\mathrm{AB}}{\mathrm{OA}}\) [by definition]

or, \(\frac{\sqrt{3}}{2}=\frac{A B}{300}\) or, 2AB = 300√3

or, AB = \(\frac{300 \sqrt{3}}{2}\) = 150√3

Again, from the triangle OCD (right-angled) we get,

sin 45° = \(\frac{\mathrm{CD}}{\mathrm{OC}}\) or, \(\frac{1}{\sqrt{2}}=\frac{C D}{O C}\)

or, OC = √2 CD

or, 300 = √2 CD

or, CD = \(\frac{300}{\sqrt{2}}\) =150√2

Hence in the first case the height of the kite is 150√3 metres and in the second case the height of the kite is 150√2 metres.

Clearly, the height of the kite is greater in the first case.

Example 9. The length of the flag at the roof of a building is 3.3 metres. From any point of road, the angles of elevation of the top and foot of the fiagpost are 50° and 45°. Calculate the height of building [Let tan 50° = 1.192]

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 9

 

Let AB be the building and AC be the flag post.

As per question, AC = 3.3 m,

∠AOB = 45° and ∠BOC = 50°

Now, from the right-angled triangle BOC, we get,

tan 50° = \(\frac{\mathrm{BC}}{\mathrm{OB}}\)

or, 1.192 = \(\frac{\mathrm{BC}}{\mathrm{OB}}\) or, 1.192 = \(\mathrm{AB}+\mathrm{AC}\)

or, 1.192 = \(\frac{\mathrm{AB}+3 \cdot 3}{\mathrm{OB}}\) or, 1.192OB = AB + 3.3………(1)

Again from the right-angled triangle AOB we get,

tan 45° = \(\frac{\mathrm{AB}}{\mathrm{OB}}\)

or, 1 = \(\frac{\mathrm{AB}}{\mathrm{OB}}\) or, OB = AB

∴ from (1) we get, 1.192AB = AB + 3.3

or, 1.192AB – AB = 3.3 or, 0.192AB = 3.3

or, AB = \(\frac{3.3}{0 \cdot 192}\) or, AB = 17.19 (approx.)

Hence the height of the building is 17.19 m (approx.).

Example 10. The length of the shadow of a three-storied building standing on the ground is found to be 60 metres more when the sun’s angle of elevation changes from 30° to 45°. Find the height of the three-storied building.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 10

 

Let AB be the three-storied building and BD and BC are the shadow of AB when the sun’s angles of elevation are 30° and 45° respectively.

Now, from the right-angled triangle ABC, we get,

tan 45° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) [by definition]

or, 1 = \(\frac{\mathrm{AB}}{\mathrm{BC}}\)

or, BC = AB………(1)

Again, from the right-angled triangle ABD, we get,

tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{\mathrm{BD}}\) or, √3AB =BD or, √3AB = BC + CD

or, √3AB = AB + 60 or, √3AB – AB = 60 or, AB (√3- 1) = 60

or, AB = \(\frac{60}{\sqrt{3}-1} \times \frac{\sqrt{3}+1}{\sqrt{3}+1}\)

or, AB = \(\frac{60(\sqrt{3}+1)}{3-1}=\frac{60(\sqrt{3}+1)}{2}\)

= 30 (1.732 + 1) = 30 x 2.732 = 81.96

Hence the height of three storied building is 81.96 metres (approx.).

Example 11. The heights of two buildings are 180 metres and 60 metres respectively. If the angle of elevation of the top of the first building from the foot of the second building is 60°. Calculate the angle of elevation of the top of the second building from the foot of the first.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 11

 

Let AB be the first building and CD be the second building.

As per the question, AB = 180 m and CD = 60 m.

Also, ∠ADB = 60° and ∠CBD = θ (let).

Now, from the right-angled triangle ABD we get,

tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)

or, √3 = \(\frac{180}{B D}\) or, BD = \(\frac{180}{\sqrt{3}}=60 {\sqrt{3}}\)

Again, from the right-angled triangle BCD, we get,

tan θ = \(\frac{\mathrm{CD}}{\mathrm{BD}}\)

or, tan θ = \(\frac{60}{60 \sqrt{3}}\) or, tan  θ = \(\frac{1}{\sqrt{3}}\)

tan θ = tan 30° or, θ = 30°

Hence the angle of elevation of the top of the second building from the foot of the first building is 30°.

Example 12. From a point on the same plane along the horizontal line passes through the foot of a tower, the angle of elevation of the top of the tower is 30° and the angle of elevation of the top of the tower is 60° at a point on the same straight line proceeding 50 metres nearer to the tower. Calculate the height of the tower.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 12

 

Let AB be the tower, from the point D the angle of elevation of the top A of the tower is 30° and from the point C, 60 metres nearer to the foot B of the tower, the angle of elevation of the top A of the tower is 60.

As per question, DC = 50 m, ∠ADB = 30° and ∠ACB = 60°.

Now, from the right-angled triangle ABC, we get,

tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) [by definition]

or, √3 = \(\frac{\mathrm{AB}}{\mathrm{BC}}\) or, BC = \(\frac{\mathrm{AB}}{\sqrt{3}}\)………(1)

Again, from the right-angled triangle ABD, we get,

tan 30° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) [by definition]

or, \(\frac{1}{\sqrt{3}}\) = \(\frac{\mathrm{AB}}{\mathrm{BD}}\) or, V3AB = BD

or, √3AB = BC + CD or, √3AB = \(\frac{\mathrm{AB}}{\sqrt{3}}\) + 50 [from (1)]

or, √3AB – \(\frac{\mathrm{AB}}{\sqrt{3}}\) = 50

or, AB\(\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)\) = 50

or, AB x \(\frac{2}{\sqrt{3}}\)

or, AB = \(\frac{50 \sqrt{3}}{2}\)

or, AB = 25√3

Hence the height of the tower is 25√3 m.

Example 13. Monu standing in the midst of a field, observes a flying bird in his north at an angle of elevation of 30° and after 2 minutes he observes the bird in his south at an angle of elevation of 60°. If the bird flies in a straight line all along at a height of 60√3 metres, find its speed in kilometre per hour.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 13

 

Let Monu standing at O observes the. flying bird in his north at the point N at an angle of elevation of 30° and in his south at the point S at an angle of elevation of 60°.

Also, let OC ⊥ NS and AB is the horizontal line.

As per question, OC = 60√3 metres,

∠AON = ∠ONC = 30° and ∠BOS =∠OSC = 60°

Now, from the right-angled triangle CON we get,

tan 30° = \(\frac{\mathrm{OC}}{\mathrm{CN}}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{60 \sqrt{3}}{\mathrm{CN}}\) or, CN = 180

Again, from the right-angled triangle COS, we get,

tan 60° = \(\frac{\mathrm{OC}}{\mathrm{CS}}\) [by definition]

or, √3 = \(\frac{\mathrm{OC}}{\mathrm{CS}}\) or, \(\sqrt{3}=\frac{60 \sqrt{3}}{C S}\) or, CS = 60.

∴ NS = CN + CS = 180 + 60 = 240

So, the speed of the bird = 240 metres per 2 minutes

= \(\frac{240}{2}\) metres per 1 minutes

= \(\frac{240 \times 60}{2 \times 1000}\) km/h

= 7.2 km per hour

Hence the speed of the bird = 7.2 km per hour.

Example 14. A vertical tower of 126 dcm height was bent at some point above the ground and it just touched the ground making an angle of 30° with the ground. Calculate at what height was the tower bent and at what distance did it meet the ground from the foot of the tower.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 14

 

Let the tower was AB and it was bent at C and just touched the ground at the point D making an angle of 30° with the ground.

∴ AC = CD and ∠BDC = 30°.

Now, from the right-angled triangle BCD, we get,

sin 30° = \(\frac{\mathrm{BC}}{\mathrm{BD}}\) [by definition]

or, \(\frac{1}{2}=\frac{A B-A C}{C D}\)

or, \(\frac{1}{2}=\frac{A B-C D}{C D}\)

or, 2AB – 2CD = CD

or, 2AB = 3CD

or, 2 x 126 = 3CD

or, CD = \(\frac{2 \times 126}{3}\) = 84

∴ AC = 84, ∴ BC = AB – AC -126 – 84 = 42.

Again, from the right-angled triangle BDC, we get,

tan 30° = \(\frac{\mathrm{BC}}{\mathrm{BD}}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{42}{B D}\) [∵ BC = 42]

or, BD = 42√3.

Hence the tower was bent at a height of 42 dcm from the ground and it met the ground at a distance of 42√3 dcm from the foot of the tower.

Example 15. Laxmi devi standing on a railway overbridge of 5√3 metres height observed the engine of the train from one side of the bridge at an angle of depression of 30°. But just after 2 seconds, she observed the engine at an angle of depression of 45° from the other side of the bridge. Find the speed of the train in metres per second.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 15

 

Let Laxmidevi standing at O observed the engine at first at A and then at B.

Also, let OC ⊥ AB.

As per question, OD = 5√3 metres,

∠EOA = ∠OAD = 30° and ∠BOF = ∠OBD = 60°

Now, from the right-angled triangle AOD, we get,

tan 30° = \(\frac{\mathrm{OD}}{\mathrm{AD}}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{5 \sqrt{3}}{A D}\) [∵ OD = 5√3m] or, AD = 15

Again, from the right-angled triangle BOD, we get,

tan 60° = \(\frac{\mathrm{OD}}{\mathrm{BD}}\)

or, \(\sqrt{3}=\frac{5 \sqrt{3}}{\mathrm{BD}}\) or, BD = 5

∴ AB = AD + BD = 15 + 5 = 20.

So, the speed of the train = 20 metres per 2 seconds

= \(\frac{20}{2}\) metres per second = 10 metres per second.

Hence the speed of the train = 10 m per second.

Example 16. A bridge is situated at right-angle to the bank of a lake. If one moves away a certain distance from the bridge along this side of the lake, the other end of the bridge is seen at an angle of 45° and if someone moves a further distance of 450 metres in the same direction, the other end is seen at an angle of 30°. Find the length of the bridge.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 16

 

Let the length of the bridge be AB. At the point bridge C the angle of elevation of the end A of the bridge is 45°.

At the point D, 450 metres away from C, the angle of elevation of end A is 30°.

As per question, CD = 450 metres, ∠ACB = 45° and ∠ADB = 30°.

Now, from the right-angled triangle ABC, we get, tan 45°= \(\frac{AB}{BC}\) [by definition]

or, 1 = \(\frac{AB}{BC}\) or, BC = AB……..(1)

Again, from the right-angled triangle ABD we get,

tan 30° = \(\frac{AB}{BD}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{A B}{B D}\)

or, √3A = BD

or, √3AB = BC + CD

or, √3AB = AB + 450 [from (1)]

or, √3AB-AB = 450 or, AB(√3 -1) = 450

or, AB = \(\frac{450}{\sqrt{3}-1}\)

or, \(\mathrm{AB}=\frac{450(\sqrt{3}+1)}{(\sqrt{3})^2-(1)^2} \quad \mathrm{AB}\)

or, \(\frac{450(\sqrt{3}+1)}{2}\)

or, AB = 225(√3+1)

Hence the length of the bridge =225(√3 +1)

Example 17. A tree of 20 metres height, stands on one side of a park and from a point on the top of the tree the angle of depression of the foot of chimney of furnace of the other side is 30° and the angle of elevation of the top of the chimney of furnace is 60°. Find the height of the chimney and the distance between the furnace and the tree.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 17

 

Let AB be the chimney and CD be the tree. The angle of depression of the point B is 30° and the angle of elevation of the point A is 60°.

As per questions, CD = 20 m.

Let CE is perpendicular to AB.

∴ ∠BCE = 30°, ∠ACE = 60°

∵ CE || BD, ∴ ∠CBD = 60°

Now, from the right-angled triangle BCD we get,

tan 30° = \(\frac{CD}{BD}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{\mathrm{CD}}{\mathrm{BD}}\)

or, BD = √3CD

= √3 x 20 [∵ CD = 20] = 20√3

∴ BD = CD = 20√3 metres.

Again, from the right-angled triangle ACE, we get,

tan 60° = \(\frac{AE}{CE}\) [by definition]

or, \(\sqrt{3}=\frac{A E}{20 \sqrt{3}}\) [∵ CE = 20√3 metres]

or, AE = 60

∵ CE ⊥ AB, ∴ BE = CD = 60 m

Now, AB = AE + BE = (60 + 20) m = 80 m.

Hence the height of the chimney = 80 metres and the distance between the brick kiln arid the house is 20√3 metres.

Example 18. If the angle of depression of two consecutive milestones on a road from an aeroplane are 60° and 30° respectively. Find the height of the aeroplane,

  1. when the two mile stones stand on the opposite side of the aeroplane
  2. when the two mile stones stand on the same side of the aeroplane.

Solution: Let O be the position of the observer.

1. The two miles are at the points A and B respectively on opposite sides of the observer.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 18-1

 

2. The two miles are at the points A and B respectively on the same, side of the observer.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 4 Application Of Trigonometric Ratios Heights And Distances Long Answer Question Example 18-2

 

OC is the height of the aeroplane.

Now, from the right-angled triangle OAC, we get,

tan 60°= \(\frac{OC}{AC}\)

or, \(\sqrt{3}=\frac{\mathrm{OC}}{\mathrm{AC}}\) or, OC = √3AC or, AC =\(\frac{\mathrm{OC}}{\sqrt{3}}\)……(1)

Also, from the right-angled triangle OBC, we get,

tan 30°= \(\frac{OC}{BC}\)

or, \(\frac{1}{\sqrt{3}}=\frac{\mathrm{OC}}{\mathrm{BC}}\) or, BC = √3OC……..(2)

Adding (1) and (2) we get,

AC + BC = \(\frac{\mathrm{OC}}{\sqrt{3}}\) + √3 OC

or, AB = OC \(\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)\)

or, 1 = OC \(\left(\frac{1+3}{\sqrt{3}}\right)\) [∵ The distance between two mile stones is 1 mile]

or, 1 = OC x \(\frac{4}{\sqrt{3}}\) or, OC = \(\frac{\sqrt{3}}{4}\)

Hence the height of the aeroplane is \(\frac{\sqrt{3}}{4}\) mile.

Again from right-angled triangle OAC we get,

tan 60° = \(\frac{OC}{AC}\) [by definition]

or, √3 = \(\frac{OC}{AC}\) or, AC = \(\frac{\mathrm{OC}}{\sqrt{3}}\)……..(1)

Also, from the right-angled triangle OBC we get,

tan 30° = \(\frac{OC}{BC}\) [by definition]

or, \(\frac{1}{\sqrt{3}}=\frac{\mathrm{OC}}{\mathrm{BC}}\) or, BC = √3OC……….(2)

Now, adding (1) and (2) we get,

AC + BC = \(\frac{\mathrm{OC}}{\sqrt{3}}\) + √3OC

AC + AB + AC = OC\(\left(\frac{1}{\sqrt{3}}+\sqrt{3}\right)\)

or, AB + 2AC = OC\(\left(\frac{1+3}{\sqrt{3}}\right)\)

or, AB + 2 x \(\frac{\mathrm{OC}}{\sqrt{3}}\) = OC x \(\frac{4}{\sqrt{3}}\)

or, AB = OC x \(\frac{4}{\sqrt{3}}\) – OC x \(\frac{2}{\sqrt{3}}\)

or, 1 = OC x\(\left(\frac{4}{\sqrt{3}}-\frac{2}{\sqrt{3}}\right)\) [The distance between two mile stones is 1 mile]

or, 1 – OC x \(\frac{4-2}{\sqrt{3}}\) or, 1 = OC x \(\frac{2}{\sqrt{3}}\)

or, OC = \(\frac{\sqrt{3}}{2}\)

Hence the height of the aeroplane is \(\frac{\sqrt{3}}{2}\) mile.

 

 

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles

What is complementary angles? 

If the sum of any two angles of a triangle be 90° or one right angle, then the angles are called complementary angles to each other.

For example, Let ABC be a right-angled triangle of which ∠ACB = θ.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Complementary Angles

 

Then, ∠BAC = 90°- θ.

So that ∠ACB + ∠BAC = θ + 90° – θ = 90°.

Hence ∠ACB and ∠BAC are complementary angles to each other.

WBBSE Solutions for Class 10 Maths

Determination of trigonometrical ratios of complementary

Trigonometric ratios of (90° – θ)

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Determination Of Trigometrical Ratios Of Complementary Angles

 

Let straight lines XOX’ and YOY’ intersect each other at right angles at O and ∠XOP = 90°-0. So, ∠POY = θ.

Let us draw PN ⊥ OX and PM ⊥ OY.

∴ PN = OM and PM = ON,

Class 10 Maths Solutions Wbbse

Now, sin (90° – θ) = \(\frac{\mathrm{PN}}{\dot{\mathrm{OP}}}\) [by the definition of sin θ]

= \(\frac{\mathrm{OM}}{\mathrm{OP}}=\frac{\text { base } .}{\text { hypotenuse }}=\cos \theta\)

Similarly, cos (90°-θ) = \(\frac{\mathrm{ON}}{\mathrm{OP}}\) [by definition] = \(\frac{\mathrm{PM}}{\mathrm{OP}}\)

= \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\sin \theta\)

tan (90°- 0) = \(\frac{\mathrm{PN}}{\mathrm{ON}}\) [ by definition ]

= \(\frac{\mathrm{OM}}{\mathrm{PM}}=\frac{\text { base }}{\text { perpendicular }}=\cot \theta\)

Also, \(\tan \left(90^{\circ}-\theta\right)=\frac{\sin \left(90^{\circ}-\theta\right)}{\cos \left(90^{\circ}-\theta\right)}=\frac{\cos \theta}{\sin \theta}=\cot \theta\)

\({cosec}\left(90^{\circ}-\theta\right)=\frac{\mathrm{OP}}{\mathrm{PN}}\) [by definition]= \(\frac{\mathrm{OP}}{\mathrm{OM}}\)

 

= \(\frac{\text { hypotenuse }}{\text { base }}=\sec \theta\)

Also, \({cosec}\left(90^{\circ}-\theta\right)=\frac{1}{\sin \left(90^{\circ}-\theta\right)}=\frac{1}{\cos \theta}=\sec \theta\)

\(\sec \left(90^{\circ}-\theta\right)=\frac{\mathrm{OP}}{\mathrm{ON}}\) [by definition]

Class 10 Maths Solutions Wbbse

= \(\frac{\mathrm{OP}}{\mathrm{PM}}=\frac{\text { hypotenuse }}{\text { perpendicular }}= {cosec} \theta\)

Also, \(\sec \left(90^{\circ}-\theta\right)=\frac{1}{\cos \left(90^{\circ}-\theta\right)}=\frac{1}{\sin \theta}={cosec} \theta\)

Again, \(\cot\left(90^{\circ}-\theta\right)=\frac{\mathrm{ON}}{\mathrm{PN}}\) [by definition]

= \(\frac{\mathrm{PM}}{\mathrm{OM}}=\frac{\text { perpendicular }}{\text { base }}=\tan \theta\)

Also, \(\cot \left(90^{\circ}-\theta\right)=\frac{\cos \left(90^{\circ}-\theta\right)}{\sin \left(90^{\circ}-\theta\right)}=\frac{\sin \theta}{\cos \theta}=\tan \theta\)

Hence We get,

  1. sin (90° – θ) = cos θ
  2. cos (90° – θ) = sinθ
  3. tan (90° – θ) = cot θ
  4. cosec (90° – θ) – sec θ
  5. sec (90° – θ) = cosec θ
  6. cot (90° – θ) = tan θ

From the above feature we see that, the trigonometric ratios of complementary angles change as per the following rules:

sin ⇔ cos
cosec ⇔ sec
tan ⇔ cot

Class 10 Maths Solutions Wbbse

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Multiple Choice Questions

Example 1. The value of (sin 43° cos 47° + cos 43° sin 47°) is

  1. 1
  2. sin 4°
  3. cos 4°
  4. none of these

Solution: sin 43° cos 47° + cos 43° sin 47°

= sin 43° cos (90° – 43°) + cos 43° sin (90° – 43°)

= sin 43° sin 43° + cos 43° cos 43°

= sin2 43° + cos2 43° =1.

∴ 1. 1 is correct.

Example 2. The value of \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\) is

  1. 0
  2. 1
  3. 2
  4. 3

Solution: \(\left(\frac{\tan 35^{\circ}}{\cot 55^{\circ}}+\frac{\cot 78^{\circ}}{\tan 12^{\circ}}\right)\)

= \(\frac{\tan 35^{\circ}}{\cot \left(90^{\circ}-35^{\circ}\right)}+\frac{\cot 78^{\circ}}{\tan \left(90^{\circ}-78^{\circ}\right)}=\frac{\tan 35^{\circ}}{\tan 35^{\circ}}+\frac{\cot 78^{\circ}}{\cot 78^{\circ}}\)

= 1 + 1 = 2

∴ 3. 2 is correct.

Example 3. The value of { cos (40° + θ) – sin (50° – θ) } is

  1. 2 cos θ
  2. 7 sin θ
  3. 0
  4. none of these

Solution: cos (40° + θ) – sin (50° – θ)

= cos (40° + θ) – sin { 90° – (40° + θ)}

= cos (40° + θ) – cos (40° + θ) = 0

∴ 2. 0 is correct

Class 10 Maths Solutions Wbbse

Example 4. ABC is a triangle, sin\(\left(\frac{B+C}{2}\right)\)

  1. sin\(\frac{\mathbf{A}}{2}\)
  2. cos\(\frac{\mathbf{A}}{2}\)
  3. sin A
  4. cos A

Solution: Since ABC is a triangle,

∴ ∠A + ∠B + ∠C = 180° or, ∠B + ∠C = 180° – ∠A

or, \(\frac{\angle \mathrm{B}+\angle \mathrm{C}}{2}=90^{\circ}-\frac{\angle \mathrm{A}}{2}\)

∴ \(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)=\cos \frac{\mathrm{A}}{2}\)

∴ 2. cos\(\frac{\mathbf{A}}{2}\) is correct.

Example 5. If A + B = 90° and tan A = \(\frac{3}{4}\), then the value of cot B is

  1. \(\frac{3}{4}\)
  2. \(\frac{4}{3}\)
  3. 3
  4. 4

Solution: Given that A + B = 90° or, A = 90° – B

∴ tan A = \(\frac{3}{4}\) ⇒ tan (90°-B) = \(\frac{3}{4}\) ⇒ cot B = \(\frac{3}{4}\)

∴ 1. \(\frac{3}{4}\) is correct.

Example 6. If sin 5θ = cos 4θ, then the value of θ is

  1. 10°
  2. 30°
  3. 45°

Solution: sin 5θ = cos 4θ

or, sin5θ = sin (90° – 4θ)

∴ 5θ = 90° – 4θ

or, 5θ + 4θ = 90°

or, 9θ = 90°

or, θ = \(\frac{90^{\circ}}{9}\) = l0°

∴ 2. 10° is correct.

Example 7. If tan 2θ = cot (θ + 15°), then the value of θ is

  1. 10°
  2. 20°
  3. 25°

Solution: tan 2θ = cot (θ + 15°)

or, cot (90° – 2θ) = cot (θ + 15°)

∴ 90° – 2θ = θ + 15°

or, – 2θ – θ = 15° – 90°

or, -3θ = – 75°

or,θ =  \(\frac{-75^{\circ}}{-3}\) = 25°

∴ 4. 25° is correct

Example 8. If A + B = 90°, then the value of sin2A + sin2B =

  1. -1
  2. 0
  3. 1
  4. \(\frac{1}{\sqrt{2}}\)

Solution: A + B = 90° ⇒ B = 90°- A.

Now, sin2A + sin2B

= sin2A + sin2 (90° – A)

= sin2A + cos2A = 1

∴ 3. 1 is correct.

Example 9. \(\frac{\sin 30^{\circ} 17^{\prime}}{\cos 59^{\circ} 43^{\prime}}\) =

  1. -1
  2. \(\frac{1}{2}\)
  3. \(\frac{1}{\sqrt{2}}\)
  4. 1

Solution: \(\frac{\sin 30^{\circ} 17^{\prime}}{\cos 59^{\circ} 43^{\prime}}=\frac{\sin 30^{\circ} 17^{\prime}}{\cos \left(90^{\circ}-30^{\circ} 17^{\prime}\right)}\)

= \(\frac{\sin 30^{\circ} 17^{\prime}}{\sin 30^{\circ} 17^{\prime}}=1\)

∴ 4. 1 is correct

Example 10. tan 1° tan 2° tan 3° ………. tan 89° =

  1. 0
  2. 1
  3. √3
  4. undefined

Solution: tan 1° tan 2° tan 3° ………. tan 89°

= tan (90° – 89°) tan (90° – 88°) tan (90° – 87°)………..tan 87° tan 88° tan 89°.

= (cot 89° tan 89°) (cot 88° tan 88°) (cot 87° tan 87°)……… (cot 44° tan 44°) tan 45°

= \(\begin{aligned}
&\left(\cot 89^{\circ} \cdot \frac{1}{\cot 89^{\circ}}\right)\left(\cot 88^{\circ} \cdot \frac{1}{\cot 88^{\circ}}\right)\left(\cot 87^{\circ} \cdot \frac{1}{\cot 87^{\circ}}\right) \\
& \cdots \cdots\left(\cot 44^{\circ} \cdot \frac{1}{\cot 44^{\circ}}\right) \tan 45^{\circ}
\end{aligned}\)

= 1. 1..1. 1. ………1 = 1

∴ 2. 1 is correct.

 

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles True Or False

Example 1. If the sum of two angles is 90°, then they are called complementary angles to each other.

Solution: True

Example 2. The value of cos 54° and sin 36° are equal.

Solution: rue

since cos 54° = cos (90° – 36°) = sin 36°.

Example 3. The simplified value of (sin 12° – cos 78°) is 1.

Solution: False

since (sin 12° – cos 78°) = Sin 12° – cos (90° – 12°) = sin 12° – sin 12° = 0.

Example 4. The values of sin 72° and cos 108° are equal.

Solutions: False

since sin- 72° = sin (180° – 108°) = sin 108°.

 

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Fill In The Blanks

Example 1. The value of (tan 15° x tan 45° x tan 60° x tan 75°) is ______

Solution: √3

since, We have, tan 15° x tan 45° x tan 60° x tan 75°

= (tan 15° x tan 75°) x tan 45° x tan 60°.

= tan 15° x cot 15° x 1 x √3 [∵ tan 75° = tan (90° – 15°) = cot 15°]

= tan 15° x \(\frac{1}{\tan 15^{\circ}}\) x √3 = 1 x √3 = √3.

The value of (tan 15° x tan 45° x tan 60° x tan 75°)  =√3.

Example 2. The value of (sin 12° x cos 18° x sec 78° x cosec 72°) is_______

Solution: 1

since, We have, sin 12° x cos 18° x sec 78° x cosec 12°

= sin 12° x cos 18° x sec (90°  – 12°) x cosec (90° – 18°)

= sin 12° x cos 18° x cosec 12° x sec 18°

= sin 12° x cos 18° x \(\frac{1}{\sin 12^{\circ}} \times \frac{1}{\cos 18^{\circ}}\) = 1 x 1 = 1.

The value of (sin 12° x cos 18° x sec 78° x cosec 72°)= 1.

Example 3. If A and B are complementary to each other, then sin A = _______

Solution: cos B

since A and B are complementary, then A + B = 90° or, A = 90° – B

⇒ sin A = sin (90°-B) = cos B.

sin A = cos B.

Example 4. cos 72° – sin 18° = ______

Solution: 0

since cos 72° – sin 18° = cos 72° – sin (90° – 12°) = cos 12° – cos 72° = 0.

cos 72° – sin 18° = 0.

 

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Short Answer Type Questions

Example 1. If tan 4θ x tan 6θ = 1 and 6θ is a positive acute angle, then find the value of θ.

Solution: Given that tan 4θ x tan 6θ = 1.

⇒ tan 4θ = \(\frac{1}{\tan 6 \theta}\) ⇒ tan 4θ = cot 6θ

⇒ tan 4θ ⇒ tan (90°-6θ)

⇒ 4θ = 90°-6θ ⇒ 4θ + 6θ = 90°

⇒ 10θ = 90°

⇒ θ = \(\frac{90^{\circ}}{10}\) ⇒ θ = 9°

Hence the value of θ is 9°

Example 2. If sec 5A = cosec (A + 36°) and 5A is a positive acute angle, then find the value of A.

Solution: Given that sec 5A = cosec (A + 36°)

or, cosec (90° – 5A) = cosec (A + 36°)

⇒ 90°- 5A = A + 36°

⇒ – 5A – A = 36°- 90°

⇒ -6A = -54°

⇒ A = \(\frac{-54^{\circ}}{-6}\)

⇒ A = 9°.

Hence the required value of A is 9°.

Example 3. Find the value of \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

Solution: \(\frac{2 \sin ^2 63^{\circ}+1+2 \sin ^2 27^{\circ}}{3 \cos ^2 17^{\circ}-2+3 \cos ^2 73^{\circ}}\)

= \(\frac{2 \sin ^2 63^{\circ}+2 \sin ^2\left(90^{\circ}-63^{\circ}\right)+1}{3 \cos ^2 17^{\circ}-2+3 \cos ^2\left(90^{\circ}-17^{\circ}\right)}=\frac{2 \sin ^2 63^{\circ}+2 \cos ^2 63^{\circ}+1}{3 \cos ^2 17^{\circ}-2+3 \sin ^2 17^{\circ}}\)

= \(\frac{2\left(\sin ^2 63^{\circ}+\cos ^2 63^{\circ}\right)+1}{3\left(\cos ^2 17^{\circ}+\sin ^2 17^{\circ}\right)-2}=\frac{2 \times 1+1}{3 \times 1-2}\)

= \(\frac{2+1}{3-2}=\frac{3}{1}=3\)

Hence the required value is 3.

Example 4. Find the value of tan 1° x tan 2° x tan 3° x……..tan 89°

Solution: tan 1° x tan 2° x tan 3° x………x tan 89°

= tan 1° x tan 2° x tan 3° x…………x tan 87° x tan 88° x tan 89°.

= tan 1° x tan 2° x tan 3° x…….x tan (90° – 3°) x tan (90° – 2°)x tan (90° – 1°)

= tan 1° x’ tan 2° x tan 3° x………..x cot 3° x cot 2° x cot 1°

= (tan 1° x cot 1°) x (tan 2° x cot 2°) x (tan 3° x cot 3°) x……….x (tan 44° x cot 44°) x tan 45°x (tan 44° x cot 44°) x tan 45°

= 1 X 1 X 1 X……….x 1 x tan 45° =  1 x 1 = 1.

Hence the required value is 1.

Example 5. Find the value of cot 17° \(\left(\cot 73^{\circ} \cos ^2 22^{\circ}+\frac{1}{\tan 73^{\circ} \sec ^2 68^{\circ}}\right)\)

Solution: cot 17° \(\left(\cot 73^{\circ} \cos ^2 22^{\circ}+\frac{1}{\tan 73^{\circ} \sec ^2 68^{\circ}}\right)\)

= cot 17° (cot 73° cos2 22° + cot 73° cos2 68°)

= cot (90° – 73°) { cot 73° cos2 22° + cot 73° cos2 (90° – 22°)}

= tan 73°. cot 73° (cos2 22° + sin2 22°)

= tan 73°. \(\frac{1}{\tan 73^{\circ}}\) x 1 = 1×1 = 1.

Hence the required value = 1.

Example 6. If sin 10θ = cos 8θ and 10θ is a positive acute angle, then find the value of tan 9θ.

Solution: Given that sin 10θ = cos 8θ and 10θ is acute.

∴ sin 10θ = sin (90° – 8θ)

⇒ 10θ = 90°-8θ

⇒10θ + 8θ = 90°

⇒ 18 θ = 90°

⇒ θ = \(\frac{90^{\circ}}{18}\)

⇒ θ = 5°

⇒ 9θ = 9×5° = 45°

⇒ tan 9θ = tan 45°

⇒ tan 9θ = 1.

Hence the value of tan 9θ = 1.

Example 7. If α + β = \(\frac{\pi}{2}\) then prove that \(\cos \alpha=\sqrt{\frac{\sin \alpha}{\cos \beta}-\sin \alpha \cos \beta}\)

Solution: LHS = cos α

RHS = \(\cos \alpha=\sqrt{\frac{\sin \alpha}{\cos \beta}-\sin \alpha \cos \beta}\)

= \(\sqrt{\frac{\sin \alpha}{\cos \left(90^{\circ}-\alpha\right)}-\sin \alpha \cos \left(90^{\circ}-\alpha\right)}\)

[∵ α + β = 90° ⇒ β = 90°-α]

= \(\sqrt{\frac{\sin \alpha}{\sin \alpha}-\sin \alpha \cdot \sin \alpha}\)

= \(\sqrt{1-\sin ^2 \alpha}=\sqrt{\cos ^2 \alpha}=\cos \alpha\)

∴ LHS = RHS. [Proved]

Example 8. If sin 17° = \(\frac{x}{y}\), then prove that sec 17°- sin 73° = \(=\frac{x^2}{y \sqrt{y^2-x^2}}\)

Solution: LHS = sec 17° – sin 73°

= \(\frac{1}{\cos 17^{\circ}}\) -sin (90° -17°)

= \(\frac{1}{\cos 17^{\circ}}\) -cos 17° = \(\frac{1-\cos ^2 17^{\circ}}{\cos 17^{\circ}}\)

= \(\frac{\sin ^2 17^{\circ}}{\sqrt{1-\sin ^2 17^{\circ}}}=\frac{\left(\frac{x}{y}\right)^2}{\sqrt{1-\left(\frac{x}{y}\right)^2}}\)

= \(\frac{\frac{x^2}{y^2}}{\sqrt{1-\frac{x^2}{y^2}}}\)

= \(\frac{\frac{x^2}{y^2}}{\sqrt{\frac{y^2-x^2}{y^2}}}\)

= \(\frac{\frac{x^2}{y^2}}{\frac{\sqrt{y^2-x^2}}{y}}\)

= \(\frac{x^2}{y^2} \times \frac{y}{\sqrt{y^2-x^2}}=\frac{x^2}{y \sqrt{y^2-x^2}}\)

Hence sec 17° – sin 73° = \(\frac{x^2}{y \sqrt{y^2-x^2}}\) [proved]

Example 9. If sin 51° = \(\frac{a}{\sqrt{a^2+b^2}}\), then find the value of tan 39°.

Solution: sin 51° = \(\frac{a}{\sqrt{a^2+b^2}}\)

or, \(\sin ^2 51^{\circ}=\frac{a^2}{a^2+b^2}\) [squaring]

or, \(\sin ^2\left(90^{\circ}-39^{\circ}\right)=\frac{a^2}{a^2+b^2}\)

or, \(\cos ^2 39^{\circ}=\frac{a^2}{a^2+b^2}\)

or, \(\sec ^2 39^{\circ}=\frac{a^2+b^2}{a^2}\)

or, \(1+\tan ^2 39^{\circ}=\frac{a^2+b^2}{a^2}\)

or, \(\tan ^2 39^{\circ}=\frac{a^2+b^2}{a^2}-1\)

or, \(\tan ^2 39^{\circ}=\frac{a^2+b^2-a^2}{a^2}\)

or, \(\tan ^2 39^{\circ}=\frac{b^2}{a^2}\)

or, \(\tan 39^{\circ}=\frac{b}{a}\)

Hence \(\tan 39^{\circ}=\frac{b}{a}\)

Example 10. If √2 sin (α-β) =1 and α, β are complementary, then find the value of α and β.

Solution: Given that √2 sin (α- β) = 1

or, sin (α- β) = \(\frac{1}{\sqrt{2}}\) = sin 45°

⇒ α – β = 45° …………(1)

Also, α + β = 90° …………(2) [∵ α, β are complementary. ]

Now, adding (1) and (2) we get,

2α = 135° or, α = \(\frac{135^{\circ}}{2}\) = 67.5°.

From (2) we get, β = 90° – α = 90° – 67.5° = 22.5°

Hence α = 67.5° and β = 22.5°.

 

Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Long Answer Type Questions

Example 1. Prove that if two angles α and β are complementary angles, then

  1. sin2α + sin2β = 1
  2. cotβ + cosβ = \(\frac{\cos \beta}{\cos \alpha}\) (1 +sinβ).
  3. \(\frac{\sec \alpha}{\cos \alpha}\) – cot2β = 1

Solutions:

1. LHS = sin2α + sin2β

= sin2α + sin2(90° – α) [∵ α +β = 90° ]

= sin2α + cos2α = 1.

Hence sin2α + cos2α = 1. [Proved]

2. LHS = cotβ + cosβ = \(\frac{\cos \beta}{\cos \beta}\) + cos β = \(\frac{\cos \beta}{\sin \left(90^{\circ}-\alpha\right)}\) + cos β

= \(\frac{\cos \beta}{\cos \alpha}+\cos \beta=\frac{\cos \beta}{\cos \alpha}(1+\cos \alpha)\)

= \(\frac{\cos \beta}{\cos \alpha}\left\{1+\cos \left(90^{\circ}-\beta\right)\right\}=\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\)

Hence cot β + cos β = \(\frac{\cos \beta}{\cos \alpha}(1+\sin \beta)\) [proved]

3. LHS = \(\frac{\sec \alpha}{\cos \alpha}\) – cot2β

= sec. \(\frac{1}{\cos \alpha}\) – cot2(90°-α) [∵ α + β = 90° ]

= sec α.sec α- tan2α = sec2α- tan2α = 1.

Example 2. Prove that sec212°- \(\frac{1}{\tan ^2 78^{\circ}}\) = 1.

Solution: sec212° – \(\frac{1}{\tan ^2 78^{\circ}}\)

= sec212° – cot278° = sec212° – cot2 (90° – 12°)

= sec212° – tan212° = 1.

Hence sec212° – \(\frac{1}{\tan ^2 78^{\circ}}\) = 1. [Proved]

Example 3. If A + B = 90°, then prove that 1 + \(\frac{\tan A}{\tan B}\)= sec2A.

Solution: A + B = 90° ⇒ B = 90°- A

∴ \(1+\frac{\tan \mathrm{A}}{\tan \mathrm{B}}=1+\frac{\tan \mathrm{A}}{\tan \left(90^{\circ}-\mathrm{A}\right)}\)

= \(1+\frac{\tan \mathrm{A}}{\cot \mathrm{A}}=1+\tan \mathrm{A} \cdot \tan \mathrm{A}\)

= 1 + tan2A = sec2A

Hence \(1+\frac{\tan \mathrm{A}}{\tan \mathrm{B}}\) = sec2A. ( Proved )

Example 4. Prove that

  1. cosec248° – tan242° = 1
  2. sec 70° sin 20° + cos 20° cosec 70° = 2.

Solutions:

1. cosec248° – tan242°

= cosec2(90° – 42°) – tan242

= sec242° – tan242° =1.

Hence cosec248° – tan242° = 1 . [ Proved ]

2. sec 70° sin 20° + cos 20° cosec 70°

= sec 70° sin (90° – 70°) + cos 20° cosec (90° – 20°)

= sec 70° cos 70° + cos 20° sec 20°

= \(\frac{1}{\cos 70^{\circ}}\). cos 70° + cos 20°. \(\frac{1}{\cos 20^{\circ}}\)

= 1 + 1=2.

Hence sec 70° sin 20° + cos 20° cosec 70° = 2. [ Proved ]

Example 5. Prove that cosec222° cot268° = sin222° + sin268° + cot268°.

Solution: cosec222° cot268° = cosec2(90° – 68°) cot268°

= sec268° cot268° = (1 + tan268°) cot268°

= cot268° + tan268° cot268°

= cot268° + tan268°. \(\frac{1}{\tan ^2 68^{\circ}}\)

= cot268° + 1

= cot268° + sin222° + cos222°

= cot268° + sin222° + cos2(90° – 68°)

= cot268° + sin222° + sin268°

= sin222° + sin268° + cot268°.

Hence cosec222° cot268° = sin222° + sin268° + cot268°. [Proved]

Example 6. Prove that cot 12° cot 38° cot 52° cot 78° cot 60°= \(\frac{1}{\sqrt{3}}\)

Solution: cot 12° cot 38° cot 52° cot 78° cot 60°

= (cot 12° cot 78°) (cot 38° cot 52°) cot 60°

= {cot 12° cot (90° – 12°)} [cot 38° cot (90° – 38°)} cot 60°.

= (cot 12° tan 12°) (cot 38° tan 38°) cot 60°.

= 1 x 1 x \(\frac{1}{\sqrt{3}}\) = \(\frac{1}{\sqrt{3}}\)

Hence cot 12 cot 38° cot 52° cot 78° cot 60° = \(\frac{1}{\sqrt{3}}\) [proved]

Example 7. ABCD Is a rectangular figure, joining A, C to prove that

  1. tan ∠ACD = cot ∠ACB
  2.  tan2 ∠CAD + 1 = \(\frac{1}{\sin ^2 \angle B A C}\)

Solutions:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Long Answer Question Example 7

 

1. tan ∠ACD

= tan (90° – ∠ACB) [∵ ∠ACD + ∠ACB = 90°]

= cot ∠ACB.

Hence tan ∠ACD = cot ∠ACB. [Proved].

2. tan2 ∠CAD + 1

= sec2 ∠CAD

= \(\frac{1}{\cos ^2 \angle \mathrm{CAD}}\)

= \(\frac{1}{\cos ^2\left(90^{\circ}-\angle \mathrm{BAC}\right)}\)

[∵ ∠CAD + ∠BAC = 90°]

= \(\frac{1}{\sin ^2 \angle \mathrm{CAD}}\)

Hence tan2 ∠CAD + 1 = \(\frac{1}{\sin ^2 \angle \mathrm{CAD}}\) [proved]

Example 8. Find the value of sin2 5° + sin2 10° + sin2 15° +…….+ sin2 90°.

Solution: sin2 5° + sin2 10° + sin2 15° + ………+ sin2 85° + sin2 90°.

= (sin2 5° + sin2 85°) + (sin2 10° + sin2 80°) + (sin2 15° + sin2 75°) + (sin2 20° + sin2 70°) + (sin2 25° + sin2 65°) + (sin2 30° + sin2 60°) + (sin2 35° + sin2 55°) + (sin2 40° + sin2 50°) + sin2 45° + sin2 90°

= {sin2 5° + sin2 (90° – 5°)} + {sin2 10° + sin2 (90° – 10°)} + {sin2 15° + sin2 (90°-15°)} + ……… + sin2 45° + sin2 90°.

= (sin2 5° + cos2 5°) + (sin2 10° + cos2 10°) + (sin2 15° + cos2 15°) ……………..+ sin2 45° + sin2 90°.

= 1 + 1 + 1 + (upto 8 terms) + \(\left(\frac{1}{\sqrt{2}}\right)^2\) + 1

= 8 + \(\frac{1}{2}\) + 1 =9 \(\frac{1}{2}\)

Hence the required value = 9 \(\frac{1}{2}\)

Example 9. AOB is a diameter of a circle with center O and C is any point on the circle, joining A, C; B, C; and O, C, prove that

  1. tan ∠ABC = cot ∠ACO
  2. sin2 ∠BCO + sin2 ∠ACO = 1.
  3. cosec2 ∠CAB – 1 = tan2 ∠ABC.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 3 Trigonometric Ratios Of Complementary Angles Long Answer Question Example 9

 

since AOB is a diameter and C is any point on the circle,

∴ ∠ACB is a semicircular angle.

∴ ∠ACB = 90°

∴ AB is a hypotenuse of the right-angled triangle ABC.

Again, since ∠ACB = 90°, ∴ ∠BAC + ∠CBA = 90°.

Now, 1. tan ∠ABC = tan (90° – ∠CAB)

= cot ∠CAB……….(1)

Again, in ΔAOC, OA = OC [∵ radii of same circle]

⇒ ∠OAC = ∠ACO

⇒ ∠CAB = ∠ACO

∴ from (1) we get, tan ∠ABC = cot ∠ACO. [ Proved ]

2. sin2 ∠BCO + sin2 ∠ACO.

= sin2 ∠BCO + sin2 ∠OAC [OC = OA, ∴ ∠ACO = ∠OAC. ]

= sin2 ∠BCO + sin2 ∠CAB

= sin2 ∠BCO + sin2 (90° – ∠ABC)

= sin2 ∠BCO + cos2 ∠ABC

= sin2∠BCO + cos2 ∠OBC

= sin2 ∠BCO + cos2 ∠BCO [OB = OC, ∠OBC =∠BCO ]

= 1.

Hence sin2 ∠BCO + sin2∠ACO = 1 [Proved]

3. cosec2 ∠CAB – 1

= cot2 ∠CAB

= cot2 (90° – ∠ABC)

= tan2 ∠ABC.

Hence cosec2 ∠CAB – 1 = tan2 ∠ABC. [ Proved ]

Example 10. If sin α – cos α = 0 find the value of \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)

Solution: \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)

= cosec α + sec α = \(=\frac{1}{\sin \alpha}+\frac{1}{\cos \alpha}=\frac{\cos \alpha+\sin \alpha}{\sin \alpha \cos \alpha}\)

= \(\frac{\sin \alpha+\sin \alpha}{\sin \alpha \sin \alpha}\) [∵ sinα – cosα= 0 ⇒  sin α= cos α]

Again, sin α = cos α = sin (90° – α)

⇒ α = 90°- α or, 2α= 90° or, α = \(\frac{90^{\circ}}{2}\) = 45°.

∴ \(\frac{\sin \alpha+\sin \alpha}{\sin \alpha \cdot \sin \alpha}\)

= \(\frac{2 \sin \alpha}{\sin ^2 \alpha}=\frac{2}{\sin \alpha}\)

= \(\frac{2}{\sin 45^{\circ}}=\frac{2}{\frac{1}{\sqrt{2}}}=2 \sqrt{2}\)

Hence \(\sec \left(\frac{\pi}{2}-\alpha\right)+ {cosec}\left(\frac{\pi}{2}-\alpha\right)\)= 2√2.

Example 11. Prove that tan 20° + tan 70° = \(\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\)

Solution: tan 20° + tan 70°

= tan 20° + tan (90° – 20°)

= tan 20° + cot 20°

= tan 20° + \(\frac{1}{\tan 20^{\circ}}=\frac{\tan ^2 20^{\circ}+1}{\tan 20^{\circ}}\)

= \(\frac{\sec ^2 20^{\circ}}{\sqrt{\tan ^2 20^{\circ}}}=\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\)

Hence tan 20° + tan 70° = \(\frac{\sec ^2 20^{\circ}}{\sqrt{\sec ^2 20^{\circ}-1}}\) [Proved]

 

 

 

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities

Definition of trigonometric ratios:

In your earlier classes you have studied about the ratios between two similar real quantities.

So, it now arises a very vital quiry to all of you that what trigonometric ratios are.

In reply, we can say that trigonometric ratios are the ratios between the lengths of any two sides of a triangle with respect to some preassigned angle of that triangle.

The triangle and its angle in respect to which the trigonometric ratios are to be determined may be of any type.

But in this class 10 as per syllabus we shall discuss only the trigonometric ratios of a right angled triangle and with respect to such an angle which lies in between the domain 0° ≤ θ ≤ 90° where θ is the angle.

WBBSE Solutions for Class 10 Maths

Now, it is clear that a triangle consists of three angles. So, we can determine at most 18 trigonometric ratios with respect to these three angles, i.e., with respect to each angle we can determine at most 6 trigonometric ratios, so far our knowledge permit us. Let us determine the ratios.

Class 10 Maths Solutions Wbbse

Let ABC be a right angled triangle of which ∠B is a right angle. So, the other two angles ∠A and ∠C are acute angles.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric

 

Here, AB ⊥ BC and it is obvious that AC is its hypotenuse. So, with respect to the ∠C. AB is the perpendicular and BC is the base.

However, if our consideration of angle be ∠A, then the opposite side of ∠A will be considered as the perpendicular, i.e., BC is the perpendicular and the adjacent side of ∠A will be considered as the base, i.e.. AB is the base with respect to the ∠A.

Let ∠C = θ (theta). Then the six trigonometric ratios with respect to θ are defined as follows:

sine of ∠C = sin θ (writing in short) = \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

cosine of ∠C = cos θ (writing in short) = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)

tangent of ∠C = tan θ (writing in short) = \(\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{AB}}{\mathrm{BC}}\)

cosecant of ∠C = cosec θ (writing in short) = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{\mathrm{AC}}{\mathrm{AB}}\)

secant of ∠C = sec θ (writing in short) = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{\mathrm{AC}}{\mathrm{BC}}\)

cotangent of ∠C = cot θ (writing in short) = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\mathrm{BC}}{\mathrm{AB}}\)

By definition, these are the six trigonometric ratios with respect to the ∠C.

Observe minutely that the trigonometric ratios cosecθ, secθ, and cotθ are clearly the inverse ratios, i.e., multiplicative opposite ratios of sinθ cosθ, and tanθrespectively.

Similarly, if the angle of consideration be ∠A = β (Beta, let), then the six trigonometric ratios are:

sin β = \(\frac{\text { perpendicular }}{\text { hypotenuse }}=\frac{\mathrm{BC}}{\mathrm{AC}}\)

cos β = \(\frac{\text { base }}{\text { hypotenuse }}=\frac{\mathrm{AB}}{\mathrm{AC}}\)

tan β = \(\frac{\text { perpendicular }}{\text { base }}=\frac{\mathrm{BC}}{\mathrm{AB}}\)

cosec β = \(\frac{\text { hypotenuse }}{\text { perpendicular }}=\frac{\mathrm{AC}}{\mathrm{BC}}\)

sec β = \(\frac{\text { hypotenuse }}{\text { base }}=\frac{\mathrm{AC}}{\mathrm{AB}}\)

cot β = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\mathrm{AB}}{\mathrm{BC}}\)

Class 10 Maths Solutions Wbbse

What do we mean by the trigonometric term sin θ?

From the definition given above, we see that the trigonometric term sinθ, it means that sinθ is purely a ratio between the two lengths of the sides of a right-angled triangle and the ratio is simply a real fraction either proper or improper.

Since sinθ is a fraction, so it holds all the properties of a fraction of real numbers.

A question now arises:

Is sinθ a product of sin and θ?

In reply we shall say that no, sinθ is not a product of sin and θ. It means sine of the angle θ, i.e., it means the ratio of perpendicular and hypotenuse of a right-angled triangle with respect to the angle θ.

Similar is the case with cosθ, tanθ, cosecθ, secθ, and tanθ.

Meaning of square of sin θ:

Square of sinθ =(sin θ)2 = sin2θ ≠ sin θ2, i.e., square of sinθ means the square of the ratio as a whole, but not the square of the angle related.

Similarly, (cosθ)2 = cos2θ, (tanθ)2 = tan2θ,

(cosecθ)2 = cosec2θ, (secθ)2 = sec2θ, (cotθ)2 = cot2θ.

\(\sqrt{\sin \theta}=\sin ^{\frac{1}{2}} \theta\) and so on.

 

\(\sqrt[3]{\sin \theta}=\sin ^{\frac{1}{3}} \theta\) and so on.

 

(sin θ)3 = sin3θ and so on.

Can the value of sinθ be greater than 1?

In a right-angled triangle, we know that the hypotenuse is always greater than the other two sides of the triangle of which one is the perpendicular and the other is the base. We also know that

\(\sin \theta=\frac{\text { perpendicular }}{\text { hypotenuse }}\)

Class 10 Maths Solutions Wbbse

Thus, in the fraction of sinθ, the denominator is always greater than the numerator unless and until θ takes the value 90°.

If θ = 90°, then the perpendicular and hypotenuse become the same, but the hypotenuse can never be greater than the perpendicular.

So, the value can never be greater than 1, but it may take the value 1 when the angle considered is 90°.

Can the value of cos0 be greater than 1?

We know that cosθ = \(\frac{\text { base }}{\text { hypotenuse }}\) and the hypotenuse is always greater than or equal to the base of a right-angled triangle.

So, the value of cosθ may be at most 1. but can never be greater than 1.

All other trigonometric ratios like tanθ, cosecθ, secθ, and cotθ may take any real value.

We can now conclude the properties of trigonometric ratios as follows:

Properties of trigonometric ratios:

  1. Trigonometric ratios are the ratios between the length of any two sides of a right-angled triangle.
  2. Trigonometric, ratios obey the properties of real-valued numerical fractions.
  3. The trigonometric ratio sinθ may take the value 1 at most, but can never be greater than 1.
  4. The trigonometric ratio cosθ may take the value 1 at most, but can never be greater than 1.
  5. The trigonometric ratios tanθ, cosecθ, secθ, and cotθ may take any real value.
  6. The trigonometric ratio sinθ is not a product of sin and θ. Similar are the cases with all other trigonometric ratios.
  7. Trigonometric ratios are measured with respect to some angle of a triangle.
  8. Only the term sin does not mean anything unless and untill an angle, say, θ is joined to it.

Relation between trigonometric ratios:

By the definition of trigonometric ratios we have seen that

Class 10 Maths Solutions Wbbse

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Relation Between Trigonometric Ratios

 

Moreover, tanθ = \(\frac{\text { perpendicular }}{\text { base }}\)=\(\frac{\text { perpendicular }}{\text { hypotenuse }}\) [ Dividing by hypotenuse]

= \(\frac{\sin \theta}{\cos \theta}\) [by definition of sin θ and cos θ]

Similarly, cotθ = \(\frac{\text { base }}{\text { perpendicular }}=\frac{\frac{\text { base }}{\text { hypotenuse }}}{\frac{\text { perpendicular }}{\text { hypotenuse }}}\)

= \(\frac{\cos \theta}{\sin \theta}\) [by definition of cosθ and sinθ]

Trigonometric ratios of some standard angles

In trigonometry, the following angles are of immense use : θ°, 30°, 45°, 60°, 90°. So these angles are known as standard angles in trigonometry.

We shall now find the trigonometric ratios of these angles in the following table:

Ratios 30° 45° 60° 90°
sin 0 \(\frac{1}{2}\) \(\frac{1}{\sqrt{2}}\) \(\frac{\sqrt{3}}{2}\) 1
cos 1 \(\frac{\sqrt{3}}{2}\) \(\frac{1}{\sqrt{2}}\) \(\frac{1}{2}\) 0
tan 0 \(\frac{1}{\sqrt{3}}\) 1 √3 undefined
cosec undefined 2 √2 \(\frac{2}{\sqrt{3}}\) 1
sec 1 \(\frac{2}{\sqrt{3}}\) √2 2 undefined
cot undefined √3 1 \(\frac{1}{\sqrt{3}}\) 0

 

The student can easily remember these values by the short-cut procedure given below:

write → 0, 1, 2, 3, 4

Divide by 4 → \(\frac{0}{4}, \frac{1}{4}, \frac{2}{4}, \frac{3}{4}, \frac{4}{4}\)

= \(0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1\)

Take square root →

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Ratios

 

All other trigonometric ratios can easily be found by the formulas given in the relation between the trigonometric ratios.

Trigonometric ratios when the angles are negative:

  1. sin(- θ) = – sin θ
  2. cos(- θ) = cos θ
  3. tan (- θ) = – tan θ
  4. cosec(- θ) = – cosec θ
  5. sec(- θ) = sec θ
  6. cot(- θ) = – cot θ

From the above formulas we see that only the cos θ and sec θ are not affected by the negative angles.

So these two trigonometric ratios are said to be even trigonometric functions and all others are known as odd trigonometric functions.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Multiple Choice Questions

Example 1. \(\frac{\frac{1}{3} \cos 30^{\circ}}{\frac{1}{2} \sin 45^{\circ}}+\frac{\tan 60^{\circ}}{\cos 30^{\circ}}\)

  1. \(\frac{6+\sqrt{6}}{\sqrt{3}}\)
  2. \(\frac{6-\sqrt{6}}{\sqrt{3}}\)
  3. \(\frac{6+\sqrt{6}}{3}\)
  4. \(\frac{6-\sqrt{6}}{3}\)

Solution: \(\frac{\frac{1}{3} \cos 30^{\circ}}{\frac{1}{2} \sin 45^{\circ}}+\frac{\tan 60^{\circ}}{\cos 30^{\circ}}\)

= \(\frac{\frac{1}{3} \times \frac{\sqrt{3}}{2}}{\frac{1}{2} \times \frac{1}{\sqrt{2}}}+\frac{\sqrt{3}}{\frac{\sqrt{3}}{2}}=\frac{\frac{1}{2 \sqrt{3}}}{\frac{1}{2 \sqrt{2}}}+\sqrt{3} \times \frac{2}{\sqrt{3}}\)

= \(\frac{1}{2 \sqrt{3}} \times \frac{2 \sqrt{2}}{1}+2=\frac{\sqrt{2}}{\sqrt{3}}+2=\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}+2\)

= \(\frac{\sqrt{6}}{3}+2=\frac{\sqrt{6}+6}{3}=\frac{6+\sqrt{6}}{3}\)

∴ 3. \(\frac{6+\sqrt{6}}{3}\) is correct.

Example 2. sin2 45° + cos245° =

  1. 1
  2. -1
  3. 0
  4. 2

Solution: sin2 45°+cos2 45° = \(\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2=\frac{1}{2}+\frac{1}{2}=1\)

∴ 1. 1 is correct.

Example 3. \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

  1. 1
  2. 0
  3. √2
  4. √3

Solution: \(\frac{2 \tan 30^{\circ}}{1-\tan ^2 30^{\circ}}\)

= \(\frac{2 \times \frac{1}{\sqrt{3}}}{1-\left(\frac{1}{\sqrt{3}}\right)^2}=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}=\frac{2}{\sqrt{3}} \times \frac{3}{2}=\sqrt{3}\)

∴ 4. √3 is correct.

Example 4. \(\tan ^2 \frac{\pi}{2} \sin \frac{\pi}{3} \tan \frac{\pi}{6} \tan ^2 \frac{\pi}{3}\)

  1. 1
  2. 1\(\frac{1}{2}\)
  3. 1\(\frac{1}{3}\)
  4. 1\(\frac{1}{4}\)

Solution: \(\tan ^2 \frac{\pi}{2} \sin \frac{\pi}{3} \tan \frac{\pi}{6} \tan ^2 \frac{\pi}{3}\)

= tan245° sin 60° tan 30° tan260° [ π = 180°]

= \((1)^2 \times \frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}} \times(\sqrt{3})^2=1 \times \frac{1}{2} \times 3=\frac{3}{2}=1 \frac{1}{2}\)

∴ 2. 1\(\frac{1}{2}\) is correct.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities True Or False

Example 1. sinθ is a product of sin and θ.

Solution: False

sinθ is a ratio between the two sides of a right-angled triangle with respect to the angle θ.

Example 2. The value of cos0 can never be greater than 1.

Solution: True

Since cos θ = \(\frac{\text { base }}{\text { hypotenuse }}\) and hypotenuse of a right-angled mangle can never be greater than its base with respect to any angle of it.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Fill In The Blanks

Example 1. \(\tan \theta=\frac{\text { perpendicular }}{}\)

Solution: Base

Example 2. sec 45° = \(\frac{1}{}\)

Solution: \(\frac{1}{\sqrt{2}} ; \quad \text { since } \quad \sec 45^{\circ}=\frac{1}{\cos 45^{\circ}}=\frac{1}{\frac{1}{\sqrt{2}}}\)

Example 3. cot 0° = _______

Solution: undefined

since cot0°= \(\frac{\cos 0^{\circ}}{\sin 0^{\circ}}=\frac{1}{0}\) = undefined.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Short Answer Type Questions

Example 1. If cosθ = 0.6, then show that (5 sinθ – 3 tanθ) = 0.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Short Answer Question Example 1

 

Example 2. Find the value of sin2 45° – cosec2 60° + sec2 30°.

Solution: sin2 45°- cosec260° + sec230°

= \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{2}{\sqrt{3}}\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2=\frac{1}{2}-\frac{4}{3}+\frac{4}{3}=\frac{1}{2}\)

Hence the required value \(\frac{1}{2}\).

Example 3. Prove that cos 60° = cos2230° – sin20°.

Solution: LHS = cos 60°

= \(\frac{1}{2}\)

RHS = cos230° – sin230°

= \(\left(\frac{\sqrt{3}}{2}\right)^2-\left(\frac{1}{2}\right)^2=\frac{3}{4}-\frac{1}{4}=\frac{3-1}{4}\)

=\(\frac{2}{4}=\frac{1}{2}\)

∴ LHS= RHS. (Proved)

Example 4. Prove that \(\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}\) = sec 60° + tan 60°

Solution: LHS \(\sqrt{\frac{1+\cos 30^{\circ}}{1-\cos 30^{\circ}}}\)

=\(\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{2}}}\)

= \(\sqrt{\frac{\frac{2+\sqrt{3}}{2}}{\frac{2-\sqrt{3}}{2}}}=\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}\)

= \(\sqrt{\frac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\)

= \(\sqrt{\frac{(2+\sqrt{3})^2}{4-3}}=\sqrt{(2+\sqrt{3})^2}=2+\sqrt{3}\)

RHS = sec 60° + tan 60°

= 2 + √3

∴ LHS = RHS. [Proved]

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Type Questions

Example 1. In the window of a house, there is a ladder at an angle of 60° with the ground. If the ladder is 2√3 m long, then calculate the height of the window above the ground.

Solution: Let the height of the ground be AB. Then from the right-angled triangle ABC. we get.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 1

 

sin 60° = \(\frac{AB}{AC}\) [ by the definition of sin θ]

or, \(\frac{\sqrt{3}}{2}\) = \(\frac{A B}{2 \sqrt{3}}\)

or, AB = 3

Hence the window is 3 m high above the ground.

Example 2. ABC is a right-angled triangle w ith its ∠B is 1 right angle. If AB = 8√3 cm and BC = 8 cm, then calculate the values of ∠ACB and ∠BAC.

Solution: In the right-angled triangle ABC, ∠B = 90°.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 2

 

∴∠ACB + ∠BAC = 90°…….(1)

Now, by the definition of tan θ,

tan ∠ACB = \(\frac{\mathrm{AB}}{\mathrm{BC}}\left[\tan \theta=\frac{\text { perpendicular }}{\text { base }}\right]\)

= \(\frac{8 \sqrt{3}}{8}\)

= √3 = tan60°

∴ ∠ACB = 60°

From (1) we get, ∠BAC = 90° – ∠ACB = 90° – 60° = 30°.

Hence the required value of ∠ACB = 60° and ∠BAC = 30°.

Example 3. In a right-angled triangle ABC, ∠B = 90°, ∠A = 30°, and AC = 20 cm. Determine the lengths of two sides BC and AB.

Solution: From the right-angled triangle ABC, cos 30°= \(\frac{AB}{AC}\)

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 3

 

or, \(\frac{\sqrt{3}}{2}=\frac{A B}{20}\)

\(2 A B=20 \sqrt{3}\) or, \(\quad \mathrm{AB}=\frac{20 \sqrt{3}}{2}=10 \sqrt{3}\)

Also in the triangle ABC, sin 30°= \(\frac{BC}{AC}\) or, \(\frac{1}{2}\) = \(\frac{BC}{20}\)

or, 2BC = 20 or, BC = \(\frac{20}{2}\) = 10

Hence the lengths of both BC is 10 cm and AB is 10√3

Example 4. In a right-angled triangle PQR, ∠Q = 90°, ∠R = 45°. If PR = 3√2 ,then find the lengths of two sides PQ and QR.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 4

 

In the right-angled triangle PQR, ∠Q = 1 right-angle and ∠R = 45°

∴ sin 45°= \(\frac{PQ}{PR}\) [ by definition ]

or, \(\frac{1}{\sqrt{2}}=\frac{P Q}{3 \sqrt{2}}\) or, PQ = 3

Again, cos 45°= \(\frac{QR}{PR}\) [by definition]

or, \(\frac{1}{\sqrt{2}}=\frac{\mathrm{QR}}{3 \sqrt{2}}\) or, QR = 3

Hence the lengths of both PQ and QR is 3 m.

Example 5. Calculate: sin245° – cosec260° + sec230°.

Solution: sin245°- cosec20° + sec20°.

= \(\left(\frac{1}{\sqrt{2}}\right)^2-\left(\frac{2}{\sqrt{3}}\right)^2+\left(\frac{2}{\sqrt{3}}\right)^2\)

= \(\frac{1}{2}\)

Example 6. Calculate: sec245° – cot245° – sin230° – sin260°.

Solution: sec245°- cot245°- sin230°- sin260°

= \((\sqrt{2})^2-(1)^2-\left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{3}}{2}\right)^2\)

= \(2-1-\frac{1}{4}-\frac{3}{4}=\frac{8-4-1-3}{4}=\frac{0}{4}= 0\)

Example 7. Calculate: 3 tan245° – sin260° – \(\frac{1}{3}\)cot230°- \(\frac{1}{8}\)sec245°.

Solution: 3 tan245° – sin260° – \(\frac{1}{3}\)cot230°- \(\frac{1}{8}\)sec245°.

= \(3 \times(1)^2-\left(\frac{\sqrt{3}}{2}\right)^2-\frac{1}{3} \times(\sqrt{3})^2-\frac{1}{8} \times(\sqrt{2})^2\)

= \(\frac{3}{4}-\frac{1}{3} \times 3-\frac{1}{8} \times 2=3-\frac{3}{4}-1-\frac{1}{4}=\frac{12-3-4-1}{4}=\frac{4}{4}=1\)

Example 8. Calculate: \(\frac{4}{3}\) cot230° + 3 sin2 60°- 2 cosec260°- \(\frac{3}{4}\) tan230°.

Solution: \(\frac{4}{3}\) cot230° + 3 sin260°- 2 cosec260°- \(\frac{3}{4}\) tan230°

= \(\frac{4}{3} \times(\sqrt{3})^2+3 \times\left(\frac{\sqrt{3}}{2}\right)^2-2 \times\left(\frac{2}{\sqrt{3}}\right)^2-\frac{3}{4} \times\left(\frac{1}{\sqrt{3}}\right)^2\)

= \(\frac{4}{3} \times 3+3 \times \frac{3}{4}-2 \times \frac{4}{3}-\frac{3}{4} \times \frac{1}{3}=4+\frac{9}{4}-\frac{8}{3}-\frac{1}{4}\)

= \(\frac{48+27-32-3}{12}=\frac{75-35}{12}=\frac{40}{12}=\frac{10}{3}=3 \frac{1}{3}\)

Example 9. Calculate : cot230° – 2 cos260°- \(\frac{3}{4}\)sec245° – 4 sin230°.

Solution: cot230° – 2 cos260°- \(\frac{3}{4}\)sec245° – 4 sin230°

= \((\sqrt{3})^2-2 \times\left(\frac{1}{2}\right)^2-\frac{3}{4} \times(\sqrt{2})^2-4 \times\left(\frac{1}{2}\right)^2\)

= \(3-2 \times \frac{1}{4}-\frac{3}{4} \times 2-4 \times \frac{1}{4}\)

= \(3-\frac{1}{2}-\frac{3}{2}-1=\frac{6-1-3-2}{2}=\frac{6-6}{2}=\frac{0}{2}=0\)

Example 10. Calculate: sec260°- cot230°- \(\frac{2 \tan 30^{\circ}{cosec} 60^{\circ}}{1+\tan ^2 30^{\circ}}\)

Solution: sec260°- cot230°- \(\frac{2 \tan 30^{\circ}{cosec} 60^{\circ}}{1+\tan ^2 30^{\circ}}\)

= \((2)^2-(\sqrt{3})^2-\frac{2 \times \frac{1}{\sqrt{3}} \times \frac{2}{\sqrt{3}}}{1+\left(\frac{1}{\sqrt{3}}\right)^2}\)

= \(4-3-\frac{\frac{4}{3}}{1+\frac{1}{3}}\)

= \(1-\frac{\frac{4}{3}}{\frac{3+1}{3}}=1-\frac{\frac{4}{3}}{\frac{4}{3}}=1-1=0\)

Example 11. Calculate: \(\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}\) + cos 60° cos 30° + sin 60° sin 30°.

Solution: \(\frac{\tan 60^{\circ}-\tan 30^{\circ}}{1+\tan 60^{\circ} \tan 30^{\circ}}\) + cos 60° cos 30° + sin 60° sin 30°

= \(\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}+\frac{1}{2} \times \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \times \frac{1}{2}=\frac{\frac{3-1}{\sqrt{3}}}{1+1}+\frac{\sqrt{3}}{4}+\frac{\sqrt{3}}{4}\)

= \(\frac{\frac{2}{\sqrt{3}}}{2}+\frac{2 \sqrt{3}}{4}=\frac{2}{\sqrt{3}} \times \frac{1}{2}+\frac{2 \sqrt{3}}{4}=\frac{1}{\sqrt{3}}+\frac{2 \sqrt{3}}{4}\)

= \(\frac{4+6}{4 \sqrt{3}}=\frac{10}{4 \sqrt{3}}=\frac{5}{2 \sqrt{3}}\)
\end{aligned}

Example 12. Calculate: \(\frac{1-\sin ^2 30^{\circ}}{1+\sin ^2 30^{\circ}} \times \frac{\cos ^2 60^{\circ}+\cos ^2 30^{\circ}}{{cosec}^2 90^{\circ}-\cot ^2 90^{\circ}}\) ÷ (sin 60° tan 30°)

Solution: \(\frac{1-\sin ^2 30^{\circ}}{1+\sin ^2 30^{\circ}} \times \frac{\cos ^2 60^{\circ}+\cos ^2 30^{\circ}}{{cosec}^2 90^{\circ}-\cot ^2 90^{\circ}}\) ÷ (sin 60° tan 30°)

= \(\frac{1-\left(\frac{1}{2}\right)^2}{1+\left(\frac{1}{2}\right)^2} \times \frac{\left(\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2}{(1)^2-0} \div\left(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}\right)\)

= \(\frac{1-\frac{1}{4}}{1+\frac{1}{4}} \times \frac{\frac{1}{4}+\frac{3}{4}}{1} \div \frac{1}{2}=\frac{\frac{3}{4}}{\frac{5}{4}} \times \frac{1}{1} \div \frac{1}{2}\)

= \(\frac{3}{4} \times \frac{4}{5} \times 1 \times \frac{2}{1}=\frac{6}{5}=1 \frac{1}{5}\)

Example 13.Prove that: \(\frac{2 \tan ^2 30^{\circ}}{1-\tan ^2 30^{\circ}}\) + sec245°-cot245° = sec 60°.

Solution: LHS = \(\frac{2 \tan ^2 30^{\circ}}{1-\tan ^2 30^{\circ}}\) + sec245°-cot245° + sec245°-cot245°

= \(\frac{2 \times\left(\frac{1}{\sqrt{3}}\right)^2}{1-\left(\frac{1}{\sqrt{3}}\right)^2}+(\sqrt{2})^2-(1)^2\)

= \(\frac{2 \times \frac{1}{3}}{1-\frac{1}{3}}+2-1\)

= \(\frac{\frac{2}{3}}{\frac{2}{3}}+1=1+1=2\)

RHS = sec 60° = 2

∴ LHS = RHS. [Proved]

Example 14. Prove that: sin\(\frac{\pi}{3}\) tan\(\frac{\pi}{6}\) + sin\(\frac{\pi}{2}\) cos \(\frac{\pi}{3}\) = 2 sin 2 \(\frac{\pi}{4}\)

Solution: LHS = sin\(\frac{\pi}{3}\) tan\(\frac{\pi}{6}\) + sin\(\frac{\pi}{2}\) cos \(\frac{\pi}{3}\) = 2 sin 2 \(\frac{\pi}{4}\)

= sin 60° tan 30° + sin 90° cos 60°

= \(\frac{\sqrt{3}}{2} \times \frac{1}{\sqrt{3}}+1 \times \frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1\)

RHS = 2 sin2 \(\frac{\pi}{4}\) = 2 x \(\left(\frac{1}{\sqrt{2}}\right)^2\) = 2 x \(\frac{1}{2}\) = 1.

∴ LHS = RHS. [Proved]

Example 15. If x sin 45° cos 45° tan 60°= tan245° – cos 60°, then determine the value of x.

Solution: x sin 45° cos 45° tan 60° = tan245° – cos 60°

or, \(x \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} \times \sqrt{3}=1-\frac{1}{2}\)

or, \(x \times \frac{\sqrt{3}}{2}=1-\frac{1}{2}\)

or, \(\quad x \times \frac{\sqrt{3}}{2}=\frac{1}{2}\)

or, \(\quad x=\frac{1}{\sqrt{3}}\)

Hence the required value of x = \(\frac{1}{\sqrt{3}}\).

Example 16. If x sin 60° cos230° = \(\frac{\tan ^2 45^{\circ} \sec 60^{\circ}}{{cosec} 60^{\circ}}\) then find the value of x.

Solution: Given that x sin 60° cos230° = \(\frac{\tan ^2 45^{\circ} \sec 60^{\circ}}{{cosec} 60^{\circ}}\)

or, \(x \times \frac{\sqrt{3}}{2} \times\left(\frac{\sqrt{3}}{2}\right)^2=\frac{(1)^2 \times 2}{\frac{2}{\sqrt{3}}}\)

or, \(x \times \frac{\sqrt{3}}{2} \times \frac{3}{4}=\frac{2 \sqrt{3}}{2}\)

or, \(\quad x=\frac{8}{3}=2 \frac{2}{3}\)

Hence the value of x = 2 \(\frac{2}{3}\)

Example 17. If x2 = sin230° + 4 cot245° – sec260°, determine the value of x.

Solution: x2 = sin230° + 4 cot245° – sec260°

or, \(x^2=\left(\frac{1}{2}\right)^2+4 \times(1)^2-(2)^2\)

or, \(x^2=\frac{1}{4}+4-4\)

or, \(x^2=\frac{1}{4}\)

or, \(x= \pm \frac{1}{2}\)

Hence the value of x = \(\pm \frac{1}{2}\)

Example 18. If x tan 30° + y cot 60° = 0 and 2x-y tan 45° = 1, then calculate the values of x and y.

Solution: Given that x tan 30° + y cot 60° = 0

or, \(x \times \frac{1}{\sqrt{3}}+y \times \frac{1}{\sqrt{3}}=0\)

or, x + y = 0 [ multiplying by √3 ] ……..(1)

Again, 2x- y tan 45° = 1 or, 2x- y x 1 = 1 or, 2x – y = 1 ……..(2)

Adding (1) and (2) we get, 3x = 1 or, x = \(\frac{1}{3}\)

∴ from (1) we get, \(\frac{1}{3}\) + y = 0

or, y = –\(\frac{1}{3}\)

Hence the values of x = \(\frac{1}{3}\) and y = –\(\frac{1}{3}\)

Example 19. If A = B= 45°, then justify

  1. sin (A + B) = sin A cos B + cos A sin B.
  2. cos (A + B) – cos A cos B – sin A sin B.

Solution:

1. Given that A = B = 45°.

Now, sin (A + B) = sin (45° + 45°) = sin 90° = 1.

Also, sin A cos B + cos A sin B = sin 45° cos 45° + cos 45° sin 45°

= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}+\frac{1}{2}=1 .\)

Hence, sin (A + B) = sin A cos B + cos A sin B. (Proved]

2. cos (A + B) = cos (45° + 45°) = cos 90° = 0

Also, cos A cos B – sin A sin B

= cos 45° cos 45° – sin 45° sin 45°

= \(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}=\frac{1}{2}-\frac{1}{2}\) = 0.

Hence cos (A + B) = cos A cos B – sin A sin B. (Proved]

Example 20. In an equilateral triangle ABC, BD is a median. Prove that tan ∠ABD = cot ∠BAD.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 20

 

Here, ∠ADB = 90°, ∴ AB = hypotenuse.

∴ tan ∠ABD = [by definition]……(1)

Also, cot ∠BAD = [ by definition ]……(2)

∴ from (1) and (2) we get, tan ∠ABD = cot ∠BAD.

Hence tan ∠ABD = cot ∠BAD. [Proved]

Example 21. In an isosceles triangle ABC, AB = AC and ∠BAC = 90°, the bisector of ∠BAC intersects the side BC at the point D. Prove that \(\frac{\sec \angle A C D}{\sin \angle C A D}\) cosec2 ∠CAD.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Long Answer Question Example 21

 

In ΔABC, AB = AC,

∴ ∠ABC = ∠ACB = 45° [∠BAC = 90° ]

AD is the bisector of ∠BAC = 90°.

∴ ∠BAD = 45° and ∠CAD = 45°

Now, \(\frac{\sec \angle A C D}{\sin \angle C A D}=\frac{\sec 45^{\circ}}{\sin 45^{\circ}}\)

[because \(\angle \mathrm{CAD}=\frac{1}{2} \angle \mathrm{BAC}=\frac{1}{2} \times 90^{\circ}=45^{\circ}\)]

= \(\frac{\sqrt{2}}{\frac{1}{\sqrt{2}}}=\sqrt{2} \times \sqrt{2}=2\)

Also, cosec2 ∠CAD = cosec245° = (√2)2 = 2

Hence \(\frac{\sec \angle \mathrm{ACD}}{\sin \angle C A D}\) = cosec2 ∠CAD. [Proved]

Example 22. Determine the value/values of θ (0° ≤ θ ≤; 90°) for which 2 cos2θ-3 cosθ + 1 = 0 will be true.

Solution: 2 cos2 θ – 3 cos θ + 1 = 0

or, 2 cos2 θ-2 cos θ – cos θ + 1 = 0

or, 2 cos θ (cos θ – 1) – 1 (cos θ – 1) = 0

or, (cos θ – 1)(2 cos θ – 1) = 0

∴ either cos θ – 1 = 0

⇒ cos θ= 1

⇒ cos θ = cos θ°

⇒ θ = 0°

or, 2 cos θ – 1 = 0

⇒ 2 cos θ=1

cos θ = \(\frac{1}{2}\) = cos 60°

Hence the values of θ are 0° and 60°.

Example 23. Prove that cot \(\frac{\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)cot\(\frac{7\pi}{8}\) =1.

Solution: LHS = cot\(\frac{\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)cot\(\frac{7\pi}{8}\)

= cot\(\frac{\pi}{8}\)cot\(\frac{7\pi}{8}\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)

= cot\(\frac{\pi}{8}\)\(\cot \left(\frac{\pi}{2}+\frac{3 \pi}{8}\right)\)cot\(\frac{3\pi}{8}\)cot\(\frac{5\pi}{8}\)

= cot\(\frac{\pi}{8}\)tan\(\frac{3\pi}{8}\)cot\(\frac{3\pi}{8}\)tan\(\frac{\pi}{8}\)

= cot\(\frac{\pi}{8}\) x tan\(\frac{\pi}{8}\) x tan\(\frac{3\pi}{8}\) x cot\(\frac{3\pi}{8}\)

= 1 x 1 cot\(\frac{\pi}{8}\) x tan\(\frac{\pi}{8}\) = cot \(\frac{\pi}{8}\) x \(\frac{1}{\cot \frac{\pi}{8}}\) = 1

= 1

= RHS [Proved]

Example 24. If tan 42° = 0.9, then find the value of cot 492°.

Solution: cot 492° = cot (90° x 5 + 42°)

= – tan 42° = – 0.9

Example 25.If sec (α- β) = 2 and sin (α + β) = \(\frac{1}{2}\), then find the least positive values of α and β.

Solution: Given that sec (α-β) = 2

⇒ sec (α- β) = sec 45°

⇒ α- β = 45°…….(1)

Again, sin(α + β) = \(\frac{1}{2}\) ⇒ sin (α +β) = sin 30°

= α + β = 30°…….(2)

But (2) is impossible, since α -β = 45°.

∴ sin (α + β) = sin 30° = sin (180° – 30°)

⇒ sin (α + β) = sin 150°

⇒ α + β = 150°……..(3)

Now. adding (1) and (3) we get, 2α = 195°

or, α = \(\frac{195^{\circ}}{2}=97 \frac{1}{2}^{\circ}\)

From (3) we get, α + 13 = 150°

or, \(97 \frac{1}{2}^{\circ}\) + 3 = 150° or, =150° -97\(\frac{1}{2}^{\circ}\) = 52\(\frac{1}{2}^{\circ}\)

Hence the least positive values of α = 97\(\frac{1}{2}^{\circ}\) and β = 52\(\frac{1}{2}^{\circ}\).

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities

An equation involving trigonometric ratios of an angle is called a trigonometric identity if it is true for all values of angles involved in it.

The basic trigonometric ratios are:

  1. sin2 θ+ cos2 θ=1
  2. sec2 θ=1+ tan2 θ
  3. cosec2 θ =1+ cot2 θ

We shall now proof these identities.

1. sin2 θ+ cos2 θ =1

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric identities

 

Let ΔABC be a right-angled triangle, in which ∠B = 90° and ∠ACB = 0. So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.

Now, in ΔABC by Pythagoras theorem we get, AB2 + BC2 = AC2.

or, \(\frac{A B^2}{A C^2}+\frac{B C^2}{A C^2}=\frac{A C^2}{A C^2}\) [Dividing both the sides by AC2]

or, \(\left(\frac{\mathrm{AB}}{\mathrm{AC}}\right)^2+\left(\frac{\mathrm{BC}}{\mathrm{AC}}\right)^2=1\)

or, \(\left(\frac{\text { perpendicular }}{\text { hypotenuse }}\right)^2+\left(\frac{\text { base }}{\text { hypotenuse }}\right)^2=1\)

or, \((\sin \theta)^2+(\cos \theta)^2=1\) [by definition of sin θ and cos θ]

or, \(\sin ^2 \theta+\cos ^2 \theta=1\)

Hence \(\sin ^2 \theta+\cos ^2 \theta=1\) [Proved]

2. sec2 θ= 1 + tan2 θ

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric identities

 

Let ΔABC be a right-angled triangle in which ∠B = right angle and ∠ACB = θ. So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.

Now, in ΔABC by pythagoras theorem we get, AC2 = AB2 + BC2

or, \(\frac{\mathrm{AC}^2}{\mathrm{BC}^2}=\frac{\mathrm{AB}^2}{\mathrm{BC}^2}+\frac{\mathrm{BC}^2}{\mathrm{BC}^2}\) [Dividing both the sides by BC2 ]

or, \(\left(\frac{A C}{B C}\right)^2=\left(\frac{A B}{B C}\right)^2+1\)

or, \(\left(\frac{\text { hypotenuse }}{\text { base }}\right)=\left(\frac{\text { perpendicular }}{\text { base }}\right)^2+1\)

or, (sec θ)2 = (tan θ)2 + 1 [by definition of sec θ and tan θ]

or, sec2 θ = tan2 θ + 1.

Hence sec2 θ=1+ tan2θ. [Proved]

3. cosec2 θ = 1 + cot2 θ

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric identities

 

Let ΔABC be a right-angled triangle in which ∠B = right angle and ∠ACB = 0.

So, with respect to θ, hypotenuse = AC, perpendicular = AB, and base = BC.

Now in ΔABC by pythagoras theorem we get, AB2 + BC2= AC2

or, \(\frac{\mathrm{AB}^2}{\mathrm{AB}^2}+\frac{\mathrm{BC}^2}{\mathrm{AB}^2}=\frac{\mathrm{AC}^2}{\mathrm{AB}^2}\) [Dividing both the sides by AB2]

or, \(1+\left(\frac{\mathrm{BC}}{\mathrm{AB}}\right)^2=\left(\frac{\mathrm{AC}}{\mathrm{AB}}\right)^2\)

[Dividing both the sides by AB2]

or, \(1+\left(\frac{\text { base }}{\text { perpendicular }}\right)^2=\left(\frac{\text { hypotenuse }}{\text { perpendicular }}\right)^2\)

or, 1 + (cot θ)2 = (cosec θ)2 [by definition of cot θ and cosec θ]

or, 1 + cot2 θ = cosec2 θ.

Hence cosec2 θ = 1 + cotθ [Proved]

We can deduce many all other trigonometric identities from these three basic identities. Such as:

sin2 θ + cos2 θ = 1 ⇒ sin2 θ=1- cos2 θ

⇒ \(\sin ^2 \theta+\cos ^2 \theta=1\)

⇒ \(\sin ^2 \theta=1-\cos ^2 \theta\)

⇒ \(\sin \theta= \pm \sqrt{1-\cos ^2 \theta}\)

Again, \(\cos ^2 \theta=1-\sin ^2 \theta \Rightarrow \cos \theta= \pm \sqrt{1-\sin ^2 \theta}\)

\(\sec ^2 \theta=1+\tan ^2 \theta\)

 

⇒ \(\sec ^2 \theta-\tan ^2 \theta=1\)

 

⇒ \((\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1\)

\(\sec ^2 \theta=1+\tan ^2 \theta\) ⇒ \(\sec \theta= \pm \sqrt{1+\tan ^2 \theta}\)

 

⇒ \(\tan ^2 \theta=\sec ^2 \theta-1 \Rightarrow \tan \theta= \pm \sqrt{\sec ^2 \theta-1}\)

Similarly cosec2θ = 1 + cot2 θ

⇒ cosec2 θ- cot2 θ = 1

⇒ (cosecθ + cot θ)(cosecθ- cotθ) = 1

Again, cosec2θ = 1 + cot2θ

⇒ cosec θ = ± \(\sqrt{1+\cot ^2 \theta}\)

Also cot2 θ = cosec22 θ-1

⇒ cot θ = ± \(=\sqrt{{cosec}^2 \theta-1}\)

Besides these trigonometric identities, the most important trigonometric formulas are:

  1. sin (A + B) = sin A cos B + cos A sin B.
  2. sin (A- B) = sin A cos B – cos A sin B.
  3. cos (A + B) = cos A cos B – sin A sin B.
  4. cos (A – B) = cos A cos B + sin A sin B.
  5. tan (A + B) = \(\frac{\tan A+\tan B}{1-\tan A \tan B}\)
  6. tan (A – B) = \(\frac{\tan A-\tan B}{1+\tan A \tan B}\)
  7. 2 sin A cos B = sin (A + B) + sin (A – B).
  8. 2 cos A sin B = sin (A + B) – sin (A – B).
  9. 2 sin A sin B = cos (A – B) – cos (A + B).
  10. 2 cos A cos B = cos (A + B) + cos (A – B).
  11. sin 2A = 2 sin A cos A = \(\frac{2 \tan A}{1+\tan ^2 A}\)
  12. cos 2A = cos2 A – sin2 A = 2 cos2 A – 1 = 1 – 2 sin2 A = \(\frac{1-\tan ^2 A}{1+\tan ^2 A}\)
  13. tan 2A = \(\frac{2 \tan A}{1-\tan ^2 A}\)
  14. sin 3A = 3 sin A – 4 sin3 A.
  15. cos 3A = 4 cos3 A – 3 cos A.
  16. tan 3A = \(\frac{3 \tan A-\tan ^3 A}{1-3 \tan ^2 A}\)

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Multiple Choice Questions

Example 1. If 3x = cosec a and \(\frac{3}{x}\) = cot α , then the value of 3 \(\left(x^2-\frac{1}{x^2}\right)\) is

  1. \(\frac{1}{27}\)
  2. \(\frac{1}{81}\)
  3. \(\frac{1}{3}\)
  4. \(\frac{1}{9}\)

Solution:

Given that 3x = cosec α

or, (3x)2 = cosec2 α [Squaring]

or, 9x2 = cosec2 α …….(1)

Again, \(\frac{3}{x}\) = cot a or, \(\frac{3}{x}^2\) = \(\cot ^2 \alpha\) [squaring]

or, \(\frac{9}{x^2}\) = cot2 α…….(2)

We know that cosec2 α – cot2 α – 1

or, \(9 x^2-\frac{9}{x^2}=1\) [from (1) and (2)]

or, \(9\left(x^2-\frac{1}{x^2}\right)=1\)

or, \(x^2-\frac{1}{x^2}=\frac{1}{9}\)

or, \(3\left(x^2-\frac{1}{x^2}\right)=3 \times \frac{1}{9}\)

or, \(3\left(x^2-\frac{1}{x^2}\right)=\frac{1}{3}\)

∴ 3. \(\frac{1}{3}\) is correct.

Example 2. If 2x = sec A and \(\frac{2}{x}\) = tan A, then the value of \(2\left(x^2-\frac{1}{x^2}\right)\)

  1. \(\frac{1}{2}\)
  2. \(\frac{1}{4}\)
  3. \(\frac{1}{8}\)
  4. \(\frac{1}{16}\)

Solution:

Given:

If 2x = sec A and \(\frac{2}{x}\) = tan A

We know that sec2 A- tan2 A = 1.

or, \((2 x)^2-\left(\frac{2}{x}\right)^2=1\)

[because \( \sec \mathrm{A}=2 x\) and \(\tan \mathrm{A}=\frac{2}{x}\)]

or, \(4 x^2-\frac{4}{x^2}\) =1

or, \(\quad 4\left(x^2-\frac{1}{x^2}\right)=1\)

or, \(4\left(x^2-\frac{1}{x^2}\right)=1\)

or, \(x^2-\frac{1}{x^2}=\frac{1}{4}\)

or, \(2\left(x^2-\frac{1}{x^2}\right)=2 \times \frac{1}{4}\)

or, \(2\left(x^2-\frac{1}{x^2}\right)=\frac{1}{2}\)

∴ 1. \(\frac{1}{2}\) is correct

Example 3. If tan α + cot α = 2, then the value of tan13 α + cot13α is

  1. 1
  2. 0
  3. 2
  4. None of these

Solution:

Given that \({tan} \propto+\frac{1}{\tan \alpha}\)=2

or, \(\frac{\tan ^2 \alpha+1}{\tan \alpha}=2\)

or, \(\tan ^2 \alpha+1=2 \tan \alpha\)

or, \(\tan ^2 \alpha-2 \tan \alpha+1=0\)

or, \((\tan \alpha-1)^2=0\)

or, \(\tan \alpha-1=0\)

or, \(\tan \alpha=1\)

∴ \(\quad \cot \alpha=\frac{1}{\tan \alpha}=\frac{1}{1}=1\)

∴ tan13 α+cot13 α = (1)13+(1)13 =1+1=2

∴ 3. 2 is correct.

Example 4. If sin θ – cos2 θ= 0 (0° < θ < 90°) and sec θ + cosec θ= x, then the value of x is

  1. 1
  2. 2
  3. √2
  4. 2√2

Solution:

Given:

If sin θ – cos2 θ= 0 (0° < θ < 90°) and sec θ + cosec θ= x,

x = sec θ + cosec θ

= \(\frac{1}{\cos \theta}+\frac{1}{\sin \theta}=\frac{\sin \theta+\cos \theta}{\sin \theta \cos \theta}\)

[sinθ-cosθ = 0

or, sin θ + cos θ- 2 sin θ cos θ= 0

or, 1-2 sin θ cos θ = 0

or, sinθcosθ = \(\frac{1}{2}a\)……..(1)

Again, sin θ + cos θ

= \(\sqrt{(\sin \theta+\cos \theta)^2}\)

= \(\sqrt{(\sin \theta-\cos \theta)^2+4 \sin \theta \cos \theta}\)

= \(\sqrt{0^2+4 \times \frac{1}{2}}\)

= \(\sqrt{0+2}=\sqrt{2}\)……(2)]

= \(\frac{\sqrt{2}}{\frac{1}{2}}\) [from (2) and (1)]

= \(2 \sqrt{2}\)

∴ 4. 2 √2 is correct

Example 5. If 2 cos 3θ = 1, then the value of 0 is

  1. 10°
  2. 15°
  3. 20°
  4. 30°

Solution: Given that 2 cos 3θ = 1

or, cos 3θ = \(\frac{1}{2}\) or, cos 3θ = cos 60°

⇒ 3θ= 60°

or, θ = \(\frac{60^{\circ}}{3}\)

or, θ = 20°

∴ 3. 20° is correct.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities True Or False

Example 1. If 0° ≤ α ≤ 90°, then the least value of (sec2 α + cos2 α) is 2.

Solutions: True

Since sec2 α + cos2 α

= (sec α – cos α )2 + 2sec α cos α

= (sec α – cos α)2 2 . \(\frac{1}{\cos \alpha}\). cos α

= (sec α – cos α)2 + 2

Now, the least value of (sec α – cos α)2 is 0.

∴ The least value of sec2 α + cos2 α = 0 + 2 = 2

Hence the given statement is true.

Example 2. The value of cos 0° x cos 1° x cos 2° x cos 3° x ……x cos 90° is 1.

Solution: False

Since the value of cos 0° x cos 1° x cos 2° x cos 3° x …….. x Cos 90°

= 1 x cos 1° x cos 2° x cos 3° x ……… x 0 [cos 0° = 1 and cos 90° = 0]

Hence the given statement is false.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Fill In The Blanks

Example 1. The value of \(\left(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\right)\) is _______

Solutions: 4

Since \(\frac{4}{\sec ^2 \theta}+\frac{1}{1+\cot ^2 \theta}+3 \sin ^2 \theta\)

= 4 cos2 θ + \(\frac{1}{{cosec}^2 \theta}\) + 3 sin2 θ

= 4 cos2 θ + sin2 θ + 3 sin2 θ

= 4 cos2 θ + 4 sin2 θ

= 4 (cos2 θ + sin2 θ)

= 4×1=4

Example 2. If sin (θ- 30° ) = \(\frac{1}{2}\) then the value, of value of cos θ is _______

Solution: \(\frac{1}{2}\)

Since sin (θ-30°) = \(\frac{1}{2}\)

⇒  sin (θ- 30° ) = sin 30°

⇒ θ- 30° = 30°

⇒ θ = 30° +30°

⇒ θ = 60°

∴ cos θ = cos 60° = \(\frac{1}{2}\)

Example 3. If cos2 θ- sin2 0 = 1 , then the value of cos4 θ – sin4 θ is ________

Solution: \(\frac{1}{2}\)

Since cos4 θ sin4 θ = (cos4 θ)2 – (sin2 θ)2

= (cos2 θ + sin2 θ) (cos2 θ – sin2 θ)

= 1 x (cos2 θ – sin2 θ)

= cos2 θ – sin2 θ.

= \(\frac{1}{2}\)

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Short Answer Type Questions

Example 1. If r cos θ = 2√3, r sinθ= 2 and 0° < θ < 90°, then determine the values of both r and θ.

Solution: r cosθ = 2√3

or, r2cos2 θ= 12 [squaring]…….(1)

rsinθ = 2

or, r2 sin2 θ = 4 [squaring]……..(2)

Adding (1) and (2) we get,

r2 cos2 θ+ r2 sin2 θ = 12 + 4

or, r2 (cos2 θ + sin2 θ) = 16

or, r2 = 16 [cos2 θ+ sin2 θ = 1]

or, r = 4

Again, \(\frac{r \cos \theta}{r \sin \theta}=\frac{2 \sqrt{3}}{2}\)

or, cot θ=√3

or, cot θ = cot 30° [0°<θ<90°]

⇒ θ = 30°.

Hence r = 4 and θ = 30°.

Example 2. If sin A + sin B = 2 where 0° ≤ A ≤ 90° and- 0° ≤ B ≤ 90°, then find the value of (cos A + cos B).

Solution: Given that 0° ≤ A < 90° and 0° ≤, B ≤ 90°

Also, sin. A + sin B = 2,

∴ Obviously, sin A = 1 and sin B = 1. [the greatest values of both sin A and sin B is 1 .]

Now, sin A = 1

⇒ sin A = sin 90°

⇒ A = 90°

Again, sin B = 1

⇒ sin B = sin 90°

⇒ B = 90°

∴ cos A + cos B = cos 90° + cos 90° = 0 + 0 = 0.

Hence the required value of cos A + cos B = 0.

Example 3. If 0° < θ < 90°, then calculate the least value of (9 tan2 θ + 4 cot2 θ).

Solution: 9 tan2 θ + 4 cot2 θ

= (3 tan θ)2 + (2 cot θ)2

= (3 tan θ – 2 cot θ)2 + 2.3 tan θ.2 cot θ

= (3 tan θ – 2 cot θ)2 + 12 tan θ \(\frac{1}{\tan \theta}\)

= (3 tan θ – 2 cot θ)2 + 12.

Now, the least value of (3 tan θ – 2 cot θ) is 0.

∴ The least of (9 tan2 θ + 4 cot2 θ) is (0 + 12) = 12.

Example 4. Calculate the value of (sin6 α  + cos6 α  + 3 sin2 α  cos6 α ).

Solution: sin6 α  + cos6 α + 3 sin2 α  cos6 α

= (sin2 α )3 + (cos2 α )3 – 3 sin2 α  cos2 α

= (sin2 α  + cos2 α )3 – 3 sin2 α  cos2 α  (sin2 α  + cos2 α ) + 3 sin2 α  cos2 α

= (1)3 – 3 sin2 α  cos2 α  1+3 sin2 α  cos2 α

= 1-3 sin2 α  cos2 α  + 3 sin2 α  cos2 α

= 1.

Hence the required value = 1.

Example 5. If cosec2 θ= 2 cot θ and 0° < θ < 90°, then determine the value of θ.

Solution: Given that cosec2 θ = 2 cot θ

or, 1 + cot2 θ – 2 cot θ = 0

or, (1)2 – 2.1.cot θ + (cot θ)2 = 0

or, (1 – cot θ)2 = 0

or, 1 – cot θ = 0

or, cot 0 = 1 = cot 45° [0° <θ < 90°]

⇒  θ= 45°.

Hence the value of θ is 45°.

 

Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Long Answer Type Questions

Example 1. If tan θ = \(\frac{3}{4}\), then show that \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{1}{2}\).

Solution: Given that tan θ = \(\frac{3}{4}\)

or, cot θ = \(\frac{4}{3}\)

or, cot 2 θ = \(\frac{16}{9}\) [squaring]

or, cosec2 θ- 1 = \(\frac{16}{9}\)

or, cosec2 θ = \(\frac{16}{9}\) + 1

or, cosec2 θ  = \(\frac{25}{9}\)

or, \(\frac{1}{\sin ^2 \theta}=\frac{25}{9}\)

or, sin2 θ= \(\frac{9}{25}\)

or, sin θ = \(\frac{3}{5}\)

Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{1-\frac{3}{5}}{1+\frac{3}{5}}}=\sqrt{\frac{\frac{2}{5}}{\frac{8}{5}}}=\sqrt{\frac{2}{8}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\)

Hence \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\) = \(\frac{1}{2}\)

Example 2. Express cosecθ and tanθ in terms of sinθ.

Solution: cosecθ = \(\frac{1}{\sin \theta}\)

and tanθ = \(\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)

Hence cosecθ = \(\frac{1}{\sin \theta}\) and tanθ = \(\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}\)

Example 3. If sec θ + tan θ= 2, determine the value of (sec θ – tan θ)

Solution: We know that sec2 θ – tan2 θ=1.

or, (sec θ + tan θ)(sec θ – tan θ) = 1

or, 2 (sec θ – tan θ) = 1 [sec θ + tan θ = 2 (given)]

or, sec θ – tanθ = \(\frac{1}{2}\)

Example 4. If cosec θ – cot θ= √2 – 1, calculate the value of (cosec θ + cot θ).

Solution: We know that cosec2 θ – cot2 θ = 1

or, (cosec θ + cot θ)(cosec θ – cot θ) = 1

or, (√2 – 1)(cosec θ- cot θ) = 1 [cosec θ + cot θ = √2 – 1]

= \(\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{\sqrt{2}+1}{(\sqrt{2})^2-(1)^2}\)

= \(\frac{\sqrt{2}+1}{2-1}=\sqrt{2}+1\)

Hence cosec θ – cot θ = √2 + 1.

Example 5. If Sin θ + cos θ = 1, then find the value of sin θ x cos θ.

Solution: Given that sin θ + cos θ = 1

or, (sin θ+ cos θ)2 = i (squaring]

or, sin2 θ + cos2 θ + 2 sin θ cos θ =1

or, 1 + 2 sin θ cos θ = 1

or, 2 sin θ cos θ = 0

or, sin θ cos θ = \(\frac{0}{2}\) = 0.

Hence the required value of sin θ x cos θ is 0.

Example 6. If tan θ + cot θ = 2, then determine the value of tan θ – cot θ.

Solution: Given that tan θ + cot θ = 2

or, tan θ + \(\frac{1}{\tan \theta}\) = 2 or, tan2 θ + 1 = 2 tan θ

or, tan2 θ – 2 tan θ + 1 = 0

or, (tan θ – 1)2 = 0

or, tan θ = 1 = tan 45°

⇒ θ = 45°

tan θ – cot θ = tan 45° – cot 45°

= 1 – 1 = 0.

Hence the required value of tan θ – cot θ = 0.

Example 7. If sin θ- cos θ = \(\frac{7}{13}\), then determine the value of (sin θ + cos θ).

Solution: Given that sin θ- cos θ = \(\frac{7}{13}\)

or, \((\sin \theta-\cos \theta)^2=\left(\frac{7}{13}\right)^2\) [squaring]

or, sin2 θ + cos2 θ- 2 sin θ cos θ = \(\frac{49}{169}\)

or, 1- 2 sin θ cos θ = \(\frac{49}{169}\) or, 2 sinθ cos θ = 1 – \(\frac{49}{169}\)

or, 2 sin θ cos θ = \(\frac{120}{169}\)……..(1)

Now, (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ

= 1 + 2 sin θ cos θ

= 1 + \(\frac{120}{169}\) [from (1)]

= \(\frac{289}{169}=\left(\frac{17}{13}\right)^2\)

sinθ + cosθ = \(\frac{17}{13}\)

Hence the required value of sin θ + cos θ = \(\frac{17}{13}\)

Example 8. If sin θ cos θ = \(\frac{1}{2}\), then calculate the value of (sin θ + cos θ).

Solution: (sin θ + cos θ)2 = sin2 θ + cos2 θ + 2 sin θ cos θ

= 1 + 2 sin θ cos θ

= 1 + 2 x \(\frac{1}{2}\) [sin θ cos θ = \(\frac{1}{2}\)]

= 1 + 1=2.

∴ sin θ + cos θ = √2

Aliter : sin θ cos θ = \(\frac{1}{2}\)

or, 2 sin θ cos θ = 2 x \(\frac{1}{2}\)

or, sin 2θ = 1 [2 sin θ cos θ = sin 2θ]

or, sin 2θ = sin 90° [sin 90°= 1]

⇒ 2θ = 90° ⇒ \(\frac{90^{\circ}}{2}\) 45°

∴ sin θ + cos θ = sin 45°+ cos 45°

= \(=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\frac{1+1}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)

Hence, sin θ + cos θ = √2 .

Example 9. If sec θ- tan θ = \(\frac{1}{\sqrt{3}}\), then determine the values of both sec θ and tan θ.

Solution: Given that sec θ- tan θ = \(\frac{1}{\sqrt{3}}\)……(1)

Also, sec2 θ – tan2 θ = 1

or, (sec θ + tan θ)(sec θ- tan θ) = 1

or, (sec θ + tan θ) x \(\frac{1}{\sqrt{3}}\) = 1 [from (1)]

or, sec θ + tan θ = √3 ………(2)

Now, adding (1) and (2) we get, 2secθ = \(\frac{1}{\sqrt{3}}\) + √3

or, 2sec θ = \(\frac{1+3}{\sqrt{3}}\)

or, 2 sec θ = \(2 \sec \theta=\frac{4}{\sqrt{3}}\)

or, sec θ = \(2 \sec \theta=\frac{2}{\sqrt{3}}\) = sec 30°

⇒ θ  = 30°

∴ tan θ = tan 30° = \(\frac{1}{\sqrt{3}}\)

Hence sec θ = \(\frac{2}{\sqrt{3}}\) and tan 0 = \(\frac{1}{\sqrt{3}}\).

Example 10. If cosec θ + cot θ = √3, then determine the value of both cosec θ and cot θ.

Solution: We know that cosec2 θ – cot2 θ = 1

or, (cosec θ + cot θ)(cosec θ – cot θ) = 1

or, √3 (cosec θ – cot θ) = 1

or, cosec θ – cot θ = \(\frac{1}{\sqrt{3}}\)……(1)

Also given that cosec θ + cot θ = √3 ………(2)

Now, adding (1) and (2) we get,

2 cosec θ = \(\frac{1}{\sqrt{3}}\) + √3

or, 2 cosec θ = \(\frac{1+3}{\sqrt{3}}\) or, 2 cosec θ = \(\frac{4}{\sqrt{3}}\)

or, cosec θ = \(\frac{2}{\sqrt{3}}\) = cosec 60°

⇒ θ = 60°

∴ cot θ= cot 60° = \(\frac{1}{\sqrt{3}}\)

Hence cosec θ = \(\frac{2}{\sqrt{3}}\) and cot θ = \(\frac{1}{\sqrt{3}}\)

Example 11. If \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) = 7 , then find the value of tan θ.

Solution: Given that \(\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}\) = 7

or, \(\frac{\sin \theta+\cos \theta+\sin \theta-\cos \theta}{\sin \theta+\cos \theta-\sin \theta+\cos \theta}=\frac{7+1}{7-1}\)

[By componerido-dividendo process]

or, \(\frac{2 \sin \theta}{2 \cos \theta}=\frac{8}{6}\)

or, tan θ = \(\frac{4}{3}\)

Hence the value of tan θ = \(\frac{4}{3}\).

Example 12. If \(\frac{{cosec} \theta+\sin \theta}{{cosec} \theta-\sin \theta}\) = \(\frac{5}{3}\), calculate the valur of sin θ.

Solution: Given that \(\frac{{cosec} \theta+\sin \theta}{{cosec} \theta-\sin \theta}= [latex]\frac{5}{3}\)

or, \(\frac{{cosec} \theta+\sin \theta-{cosec} \theta+\sin \theta}{{cosec} \theta+\sin \theta+{cosec} \theta-\sin \theta}=\frac{5-3}{5+3}\)

[By dividendo- componendo process]

or, \(\frac{2 \sin \theta}{2{cosec} \theta}=\frac{2}{8} \quad \text { or, } \quad \frac{\sin \theta}{\frac{1}{\sin \theta}}=\frac{1}{4}\)

or, \(\sin ^2 \theta=\frac{1}{4} \quad \text { or, } \sin \theta=\sqrt{\frac{1}{4}}=\frac{1}{2}\).

Hence the value of sin θ = \(\frac{1}{2}\)

Example 13. If tan2 θ + cot2 θ = \(\frac{10}{3}\), then determine the value of (tan θ + cot θ) and (tan θ – cot θ). Also find the value of tan θ.

Solution: Given that tan2 θ + cot2 θ = \(\frac{10}{3}\)

or, (tan θ)2 + (cot θ)2 = \(\frac{10}{3}\)

or, (tanθ + cotθ)2 – 2 tan θ cot θ = \(\frac{10}{3}\)

or, (tan θ + cot θ)2 – 2 \(\tan \theta \cdot \frac{1}{\tan \theta}\) = \(\frac{10}{3}\)

or, (tan θ + cot θ)2 – 2 = \(\frac{10}{3}\)

or, (tan θ + cot θ)2 = \(\frac{10}{3}\) + 2

or, \((\tan \theta+\cot \theta)^2=\frac{10+6}{3}\)

or, (tan θ+ cot θ)2 = \(\frac{16}{3}\)

or, tan θ+ cot θ = \(\frac{4}{\sqrt{3}}\)…….(1)

or, \(\tan \theta+\cot \theta=\frac{\sqrt{16}}{3}\)

Again, \(\tan ^2 \theta+\cot ^2 \theta=\frac{10}{3}\)

or, \((\tan \theta-\cot \theta)^2+2 \tan \theta \cdot \cot \theta=\frac{10}{3}\)

or, \((\tan \theta-\cot \theta)^2+2 \tan \theta \frac{1}{\tan \theta}=\frac{10}{3}\)

or, \((\tan \theta-\cot \theta)^2+2=\frac{10}{3}\)

or, \((\tan \theta-\cot \theta)^2=\frac{10}{3}-2\)

or, \((\tan \theta-\cot \theta)^2=\frac{10-6}{3}\)

or, \((\tan \theta-\cot \theta)^2=\frac{4}{3}\)

or, \(\tan \theta-\cot \theta=\sqrt{\frac{4}{3}}\)

or, \(\tan \theta-\cot \theta=\frac{2}{\sqrt{3}}\)…..(2)

Also, by adding (1) and (2) we get, 2tan θ= \(\frac{4}{\sqrt{3}}+\frac{2}{\sqrt{3}}\)

or, 2 tan θ = \(\frac{6}{\sqrt{3}}\)= or, tan θ = \(\frac{6}{\sqrt{3}}\) = √3

Hence tan θ + cot θ = \(\frac{4}{\sqrt{3}}\), tan 0- cot 0 = \(\frac{2}{\sqrt{3}}\)= and tan θ = √3.

Example 14. If sec2 θ + tan2 θ = \(\frac{13}{12}\) , then calculate the value of (sec4 θ – tan4 θ)

Solution: sec4 θ – tan4 θ

= (sec2 θ)2 – (tan2 θ)2

= (sec2 θ + tan2 θ)(sec2 θ- tan2 θ)

= \(\frac{13}{12}\) x 1 [sec2 θ – tan2 θ=1]

= \(\frac{13}{12}\)

Hence the value of sec4 θ- tan4 θ = \(\frac{13}{12}\)

Example 15. In ΔPQR, ∠Q is right angle. If PR = √5 units and PQ – RQ = 1 unit, then determine the value of cos P – cos R.

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Long Answer Question Example 15

 

In ΔPQR, ∠Q = 90°

∴ PR = hypotenuse.

Now, cos P – cos R

= \(\frac{P Q}{P R}-\frac{R Q}{P R}\)

[by definition of cos P and cos R]

= \(\frac{P Q-R Q}{P R}\)

= \(\frac{1}{\sqrt{5}}\) [PQ-RQ = 1 and PR = √5]

Hence cos P- cos R = \(\frac{1}{\sqrt{5}}\)

Example 16. If sec θ + cos θ = \(\frac{5}{2}\), then calculate the value of (sec θ – cos θ).

Solution: Given that sec θ + cos θ = \(\frac{5}{2}\)

or, (sec θ + cos θ)2 = \(\left(\frac{5}{2}\right)^2\) [squaring]

or, (sec θ- cos θ) + 4 sec θ. cos θ = \(\frac{25}{4}\)

or, (sec θ- cosθ)2 +4. \(\frac{1}{\cos \theta}\) cos θ = \(\frac{25}{4}\)

or, (sec θ- cos θ)2 + 4 = \(\frac{25}{4}\)

or, (sec θ- cos θ)2 = \(\frac{25}{4}\) – 4

or, (sec θ- cos θ)2 = \(\frac{25-16}{4}\)

or, (sec θ- cos θ)2 = \(\frac{9}{4}\)

or, sec θ- cos θ = \(\sqrt{\frac{9}{4}}\) or, sec θ-cos θ = \(\frac{3}{2}\)

Hence sec θ – cos θ = \(\frac{3}{2}\)

Example 17. Determine the value of tan θ from the relation 5 sin2 θ + 4 cos2 θ = \(\frac{9}{2}\)

Solution: Given that 5 sin2 θ + 4 cos2 θ = \(\frac{9}{2}\)

or, 10 sin2 θ+ 8 cos2 θ = 9

or, 10 sin2 θ + 8(1 – sin2 θ) = 9

or, 10 sin2 θ + 8- sin2 θ = 9

or, 2 sin2 θ = 1

or, sin2 θ = \(\frac{1}{2}\)

or, sin θ = \(\frac{1}{\sqrt{2}}\) = sin 45°

 θ = 45

∴ tan 2 θ = tan 45° = 1

Example 18. In ΔXYZ, ∠Y is right angle. If XY – 2√3 units and XZ – YZ = 2 units, then determine the value of (sec X – tan X).

Solution:

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 2 Trigonometric Ratios And Trigonometric Identities Trigonometric Identities Long Answer Question Example 28

 

In ΔXYZ, ∠Y = 90°.

∴ XZ = hypotenuse.

Now, sec X – tan X

= \(\frac{X Z}{X Y}-\frac{Y Z}{X Y}\)

[ by definition of sec X and tan X ]

= \(\frac{X Z-Y Z}{X Y}\)

= \(\frac{2}{2 \sqrt{3}}\) [XZ – YZ = 2 and XY = 2√3]

= \(\frac{1}{\sqrt{3}}\)

Hence sec X- tan X = \(\frac{1}{\sqrt{3}}\)

Example 19. Eliminate θ: x = 2 sin θ, y = 3 cos θ.

Solution: Given that x = 2 sin θ

or, \(\frac{x}{2}\) = sin θ

or, \(\frac{x^2}{4}=\sin ^2 \theta\)…….(1)

Also, y = 3 cos θ

or, \(\frac{y}{3}\) = cos θ

or, \(\frac{y^2}{9}=\cos ^2 \theta\)…….(2)

Now, adding (1) and (2) we get,

\(\frac{x^2}{4}+\frac{y^2}{9}=\sin ^2 \theta+\cos ^2 \theta\)

or, \(\frac{x^2}{4}+\frac{y^2}{9}=1\)

Hence eliminating 0 we get, \(\frac{x^2}{4}+\frac{y^2}{9}\) = 1.

Example 20. Eliminate θ: 5x = 3 sec θ, y = 3 tan θ.

Solution: Given that 5x = 3 sec 0

or, \(\frac{5x}{3}\) = sec θ

or, \(\frac{25 x^2}{9}\) = sec θ…….(1)

Again, y = 3 tan θ

or, \(\frac{y}{3}\) = tan θ

or, \(\frac{y^2}{9}=\tan ^2 \theta\)……..(2)

We know that sec2 θ – tan2 θ = 1

or, \(\frac{25 x^2}{9}-\frac{y^2}{9}=1\)

or, \(\frac{25 x^2-y^2}{9}=1\)

or, \(25 x^2-y^2=9\)

Hence eliminating θ, we get \(25 x^2-y^2=9\).

Example 21. If sin α = \(\frac{5}{13}\), then show that tan α + sec α = 1.5

Solution: sin α = \(\frac{5}{13}\)

∴ cos α = \(\sqrt{1-\sin ^2 \alpha}\)

= \(\sqrt{1-\left(\frac{5}{13}\right)^2}=\sqrt{1-\frac{25}{169}}\)

= \(\sqrt{\frac{169-25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}\)

Now, tan α + sec α = \(\frac{\sin \alpha}{\cos \alpha}+\frac{1}{\cos \alpha}\)

= \(\frac{\frac{5}{13}}{\frac{12}{13}}+\frac{1}{\frac{12}{13}}=\frac{5}{13} \times \frac{13}{12}+\frac{13}{12}\)

= \(\frac{5}{12}+\frac{13}{12}=\frac{5+13}{12}\)

= \(\frac{18}{12}=\frac{3}{2}=1 \cdot 5\)

Hence tan α + sec α = 1.5. [Proved]

Example 22. If tan A=\(\frac{m}{n}\), then determine the values of both sin A and sec A.

Solution: Given that tan A = \(\frac{m}{n}\)

or, \(\tan ^2 \mathrm{~A}=\frac{n^2}{m^2}\) [squaring]

or, \(1+\tan ^2 \mathrm{~A}=1+\frac{n^2}{m^2}\)

or, \(\sec ^2 \mathrm{~A}=\frac{m^2+n^2}{m^2}\)

or, \(\sec \mathrm{A}=\sqrt{\frac{m^2+n^2}{m^2}}\)

or, \(\sec \mathrm{A}=\frac{\sqrt{m^2+n^2}}{m}\)

tan A = \(\frac{m}{n}\)

or, cot A = \(\frac{m}{n}\) or, cot2 A = \(\frac{m^2}{n^2}\) [squaring]

or, \({cosec}^2 \mathrm{~A}-1=\frac{m^2}{n^2}\)

or, \({cosec}^2 \mathrm{~A}=\frac{m^2}{n^2}+1\)

or, \({cosec}^2 \mathrm{~A}=\frac{m^2+n^2}{n^2}\)

or, \({cosec} \mathrm{A}=\sqrt{\frac{m^2+n^2}{n^2}}\)

or, \(\frac{1}{\sin \mathrm{A}}=\sqrt{\frac{m^2+n^2}{n}}\)

or, \(\sin \mathrm{A}=\frac{n}{\sqrt{m^2+n^2}}\)

Hence \(\text { sin } \mathrm{A}=\frac{n}{\sqrt{m^2+n^2}} \text { and } \sec \mathrm{A}=\frac{\sqrt{m^2+n^2}}{m} .\)

Example 23. If cos θ = \(\frac{x}{\sqrt{x^2+y^2}}\), then show that x sin θ = y cos θ.

Solution: Given that cos θ = \(\frac{x}{\sqrt{x^2+y^2}}\)

or, cos θ = x2 [squaring] or, sec2θ = x2 +y

or, \(\cos ^2 \theta=\frac{x^2}{x^2+y^2}\) [squaring]

or, \(\sec ^2 \theta=\frac{x^2+y^2}{x^2}\)

or, \(1+\tan ^2 \theta=\frac{x^2+y^2}{x^2}\)

or, \(\tan ^2 \theta=\frac{x^2+y^2}{x^2}-1\)

or, \(\tan ^2 \theta=\frac{x^2+y^2-x^2}{x^2}\)

or, \(\tan ^2 \theta=\frac{y^2}{x^2}\)

or, \(\tan \theta=\frac{y}{x}\)

or, \(\frac{\sin \theta}{\cos \theta}=\frac{y}{x}\)

or, \(x \sin \theta=y \cos \theta\)

Hence \(x \sin \theta=y \cos \theta\) [Proved]

Example 24. If \(\sin \alpha=\frac{a^2-b^2}{a^2+b^2}\), then show that \(\cot \alpha=\frac{2 a b}{a^2-b^2}\)

Solution: Given that \(\sin \alpha=\frac{a^2-b^2}{a^2+b^2}\)

or, \(\sin ^2 \alpha=\frac{\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)^2}\) (squaring)

or, \({cosec}^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}\)

or, \(1+\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}\)

or, \(\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2}{\left(a^2-b^2\right)^2}-1\)

or, \(\cot ^2 \alpha=\frac{\left(a^2+b^2\right)^2-\left(a^2-b^2\right)^2}{\left(a^2-b^2\right)^2}\)

or, \(\cot ^2 \alpha=\frac{4 a^2 b^2}{\left(a^2-b^2\right)^2}\)

or, \((\cot \alpha)^2=\left(\frac{2 a b}{a^2-b^2}\right)^2\)

⇒ \(\cot \alpha=\frac{2 a b}{a^2-b^2}\)

Example 25. If \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\), then show that \(\sin \theta-\cos \theta=\frac{x-y}{\sqrt{x^2+y^2}}\).

Solution: Given that \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\) = k(let)

∴ sin θ = kx………(1) and cos θ = ky……..(2)

We know that sin2θ + cos2θ = 1

⇒ (kx)2 + (ky)2 = 1 [from (1) and (2)]

or, k2x2 + k2y2 = 1 or, k2(x2+y2)= 1

or, \(k^2=\frac{1}{x^2+y^2}\)

or, \(k=\sqrt{\frac{1}{x^2+y^2}}\)

or, \(k=\frac{\cdot 1}{\sqrt{x^2+y^2}}\)

∴ \(\quad \sin \theta=k x=\frac{1}{\sqrt{x^2+y^2}} \times x=\frac{x}{\sqrt{x^2+y^2}}\)……..(3)

and \(\cos \theta=k y=\frac{1}{\sqrt{x^2+y^2}} \times y=\frac{y}{\sqrt{x^2+y^2}}\)……(4)

∴ \(\sin \theta-\cos \theta=\frac{x}{\sqrt{x^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}\) [from (3) and (4)]

= \(\frac{x-y}{\sqrt{x^2+y^2}}\)

Hence \(\sin \theta-\cos \theta=\frac{x-y}{\sqrt{x^2+y^2}}\) [Proved]

Aliter: Given that \(\frac{\sin \theta}{x}=\frac{\cos \theta}{y}\)

or, \(\frac{\sin \theta}{\cos \theta}=\frac{x}{y}\)

or, \(\tan \theta=\frac{x}{y}\)

or, \(\tan ^2 \theta=\frac{x^2}{y^2}\)

or, \(\sec ^2 \theta-1,=\frac{x^2}{y^2}\)

or, \(\sec ^2 \theta=\frac{x^2}{y^2}+1=\frac{x^2+y^2}{y^2}\)

or, \(\sec \theta=\sqrt{\frac{x^2+y^2}{y^2}}\)

or, \(\sec \theta=\frac{\sqrt{x^2+y^2}}{y}\)

or, \(\cos \theta=\frac{y}{\sqrt{x^2+y^2}}\)……..(1)

∴ \(\sin \theta =\sqrt{1-\cos ^2 \theta}=\sqrt{1-\left(\frac{y}{\sqrt{x^2+y^2}}\right)^2}\)

= \(\sqrt{1-\frac{y^2}{x^2+y^2}}=\sqrt{\frac{x^2+y^2-y^2}{x^2+y^2}}=\sqrt{\frac{x^2}{x^2+y^2}}\)

= \(\frac{x}{\sqrt{x^2+y^2}}\)……….(2)

Now, \(\sin \theta-\cos \theta =\frac{x}{\sqrt{x^2+y^2}}-\frac{y}{\sqrt{x^2+y^2}}\)[from (2) and (1)]

= \(\frac{x-y}{\sqrt{x^2+y^2}}\)

Hence \(\sin \theta-\cos \theta\) = \(\frac{x-y}{\sqrt{x^2+y^2}}\) [Proved]

Example 26. If (1 + 4x2) cos A = Ax, then show that cosec A + cot A = \(\frac{1+2 x}{1-2 x}\)

Solution: Given that (1 + 4x2) cos A = 4x

or, \(\cos A=\frac{4 x}{1+4 x^2}\)

∴ sin A = \(\sqrt{1-\cos ^2 A}=\sqrt{1-\left(\frac{4 x}{1+4 x^2}\right)^2}\)

= \(\sqrt{1-\frac{16 x^2}{\left(1+4 x^2\right)^2}}\)

= \(\sqrt{\frac{\left(1+4 x^2\right)^2-16 x^2}{\left(1+4 x^2\right)^2}}\)

= \(\sqrt{\frac{1+8 x^2+16 x^4-16 x^2}{\left(1+4 x^2\right)^2}}\)

= \(\sqrt{\frac{1-8 x^2+16 x^4}{\left(1+4 x^2\right)^2}}\)

= \(\sqrt{\frac{\left(1-4 x^2\right)^2}{\left(1+4 x^2\right)^2}}\)

= \(\frac{1-4 x^2}{1+4 x^2}\)

∴ cosec A + cot A = \(\frac{1}{\sin A}+\frac{\cos A}{\sin A}=\frac{1+\cos A}{\sin A}\)

= \(\frac{1+\frac{4 x}{1+4 x^2}}{\frac{1-4 x^2}{1+4 x^2}}\)

= \(\frac{\frac{1+4 x^2+4 x}{1+4 x^2}}{\frac{1-4 x^2}{1+4 x^2}}\)

= \(\frac{1+2 \cdot 1 \cdot 2 x+(2 x)^2}{1-4 x^2}\)

= \(\frac{(1+2 x)^2}{(1+2 x)(1-2 x)}=\frac{1+2 x}{1-2 x}\)

Hence cosec A + cot A = \(\frac{1+2 x}{1-2 x}\)

Example 27. If x = a sin θ and y = b tan θ, then prove that \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\)

Solution: Given that x = a sin θ

⇒ sin θ = \(\frac{x}{a}\)

⇒ cosec θ = \(\frac{a}{x}\)

⇒ \({cosec}^2 \theta=\frac{a^2}{x^2}\)……..(1)

And y = b tan θ ⇒ tan θ = \(\frac{y}{b}\)

⇒ cot θ = \(\frac{b}{y}\)

⇒ \(\cot ^2 \theta=\frac{b^2}{y^2}\)……..(2)

We know that cosec2θ – cot2θ=1

or, \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\) = 1 [from (1) and (2)]

Hence \(\frac{a^2}{x^2}-\frac{b^2}{y^2}=1\) = 1. [Proved]

Example 28. If sin θ + sin2 θ=1, then prove that cos2 θ + cos4 θ=1.

Solution: Given that sin θ + sin2 θ=1.

or, sin θ = 1 – sin2 θ

or, sin2 θ = cos2 θ

or, sin2 θ = cos4 θ [squaring]

or, 1 – cos2 θ = cos4 θ

or, 1 = cos2 θ + cos4 θ or, cos2 θ + cos4 θ= 1

Hence cos2 θ + cos4 θ=1. [Proved]

Example 29. If 3 cot θ = 4, then find the value of \(\frac{5 \sin \theta+3 \cos \theta}{5 \sin \theta-3 \cos \theta}\)

Solution: Given that 3 cot θ = 4

or, cot θ = \(\frac{4}{3}\)

or, tan θ = \(\frac{3}{4}\)

Now, \(\frac{5 \sin \theta+3 \cos \theta}{5 \sin \theta-3 \cos \theta}=\frac{5 \cdot \frac{\sin \theta}{\cos \theta}+3}{5 \cdot \frac{\sin \theta}{\cos \theta}-3}\) [Dividing both numerator and denominator by cos θ]

= \(=\frac{5 \tan \theta+3}{5 \tan \theta-3}=\frac{5 \times \frac{3}{4}+3}{5 \times \frac{3}{4}-3}=\frac{\frac{27}{4}}{\frac{3}{4}}=9\)

Hence the required value is 9.

Example 30. If 3 tan θ = 4, then find the value of \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}\)

Solution:

Given that 3 tan θ = 4

or, tan θ = \(\frac{3}{4}\)

or, tan2 θ= \(\frac{16}{9}\)

or, sec2 θ-1 = \(\frac{16}{9}\)

or, sec2 θ = \(\frac{25}{9}\)

or sec θ = \(\frac{5}{3}\) or, cos θ = \(\frac{3}{5}\)

∴ sin θ = \(\sqrt{1-\cos ^2 \theta}\)

= \(\sqrt{1-\left(\frac{3}{5}\right)^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}\)

Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{(1-\sin \theta)(1+\sin \theta)}{(1+\sin \theta)(1+\sin \theta)}}=\sqrt{\frac{1-\sin ^2 \theta}{(1+\sin \theta)^2}}\)

= \(\sqrt{\frac{\cos ^2 \theta}{(1+\sin \theta)^2}}=\frac{\cos \theta}{1+\sin \theta}=\frac{\frac{3}{5}}{1+\frac{4}{5}}\)

= \(\frac{\frac{3}{5}}{\frac{9}{5}}=\frac{3}{9}=\frac{1}{3}\)

[Aliter: Given that 3 tan θ = 4 or, tanθ = \(\frac{4}{3}\)

Now, \(\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} =\sqrt{\frac{(1-\sin \theta)^2}{(1+\sin \theta)(1-\sin \theta)}}=\sqrt{\frac{(1-\sin \theta)^2}{1-\sin ^2 \theta}}\)

= \(\sqrt{\frac{(1-\sin \theta)^2}{\cos ^2 \theta}}=\frac{1-\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}\)

= \(\sec \theta-\tan \theta=\sqrt{1+\tan ^2 \theta}-\tan \theta\)

= \(\sqrt{1+\left(\frac{4}{3}\right)^2-\frac{4}{3}=\sqrt{\frac{25}{9}}-\frac{4}{3}}\)

= \(\frac{5}{3}-\frac{4}{3}=\frac{1}{3}\)]

Example 31. If sec θ + tan θ = p, then find the value of cos θ.

Solution: Given that sec θ + tan θ = p ….(1)

We know that sec2 θ – tan2 θ = 1

or, (sec θ + tan θ)(sec θ – tan θ) = 1

or, p (sec θ – tan θ) = 1

or, sec θ – tan θ = \(\frac{1}{p}\)…….(2)

Adding (1) and (2) we get, 2 sec θ = p + \(\frac{1}{p}\)

or, 2 sec θ = \(\frac{p^2+1}{p}\) or, sec θ = \(\frac{p^2+1}{2p}\) or, cos θ= \(\frac{2p}{p^2+1}\)

Hence cos θ = \(\frac{2p}{p^2+1}\)

Example 32. If 7 cos2 θ + 3 sin2 θ = 4 and 0° < θ < 90°, then what is the value of tan θ?

Solution: Given that 7 cos2 θ + 3 sin2 θ = 4

or, 7(1 – sin2 θ) + 3 sin2 θ = 4

or, 7-7 sin2 θ + 3 sin2 θ = 4

or, 7 – 4 sin2 θ = 4 or, -4 sin2 θ = 4-7

or, – 4 sin2 θ =- 3

or, \(\sin ^2 \theta=\frac{3}{4}\)

or, sin θ = \(\frac{\sqrt{3}}{2}\) [0°<θ<90°]

or, sin θ = sin 60°

⇒ θ = 60°

∴ tan θ = tan 60° = √3.

 

 

 

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles

Trigonometry Chapter 1 Concept Of Measurement Of Angles

What is trigonometry?

The word trigonometry is derived from the Greek words “tri” which means three, “gon” which means sides, and “metron” which means measure.

Thus trigonometry is the study of relationships between the sides and angles of a triangle.

About 2000 years ago, the famous Greek astronomer Hipparchus gave the name Trigonometry to this branch of mathematics.

This subject has been dealt with briefly in the ancient Hindu books “Surya Siddhanta” and “Poulis Siddhanta”.

In practice, in ancient times it was unknown or very difficult to measure the height of the top of a hill or the breadth of a vast river. But at present, this can be easily done with the help of trigonometry.

WBBSE Solutions for Class 10 Maths

At present, trigonometry is a great useful, and essential branch in mathematics as well as in our daily life.

Geometric and trigonometric angles: There are two types of angles.

  1. Geometrical angle and
  2. Trigonometrical angle.

1. Geometrical angle: In geometry when two lines intersect each other, angles are formed. These angles are confined in magnitude between 0° and 360°.

Class 10 Maths Solutions Wbbse

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Geometric Angles

 

In the following, two straight lines AB and PQ intersect each other at O.

So, ∠AOQ, and ∠BOP have been produced and all these angles are geometric angles.

The most important criteria of geometric angles are that

  1. They are confined in magnitude between 0° and 360° and
  2. These are all positive angles.

2. Trigonometric angle: Trigonometric angles are produced by the rotation of a ray being fixed at one of its end-point. The ray may rotate clockwise or anti-clockwise.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Trigonometric Angle

 

Let us suppose that the ray OA being fixed at the end point O rotates anti-clockwise.

Then the angles ∠AOP1, ∠P1 OP2, ∠P2OP3, ∠P3OP4 are produced.

All these are trigonometric angles.

The most important criterias of trigonometric angles are

  1. it starts from 0° and ends at so far as the ray rotates;
  2. these angles may positive or negative.

Positive and negative angles

Class 10 Maths Solutions Wbbse

Earlier you have known that geometric angles are all positive. So, positive and negative angles arise only in the case of trigonometric angles.

In trigonometric angles, if the rotating ray rotates clockwise, then negative angles are produced, but if the ray rotates anti-clockwise, then positive angles are evolved.

∠AOB is a positive angle, since here the initial line OA rotates anti-clock wise, whereas the ∠POQ is a negative angle, since here the initial line OP rotates clockwise.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Positive Angle

 

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Negative Angle

 

In trigonometric angles, one complete revolution of the ray produces 360°. Thus if a ray rotates one complete revolution and then rotates 30° more anti-clockwise, then the total angle produced = 360° + 30°= 390°.

Similar is the case for any other revolution of the ray, obviously more than one complete revolution.

Thus in trigonometric angles it is very possible to get an angle (both positive and negative) greater than 360°.

Measurement of angles

There are three systems of units for measurement of angles,

  1. Sexagesimal system
  2. Centesimal system
  3. Circular or radian system

1. Sexagesimal system:

In sexagesimal system of measurement of angles, 1 right angle is taken as a unit.

In this system, 1 right angle is divided into 90 equal parts and each part is called a degree, which is denoted by (°).

A degree is divided into 60 equal parts and each part is called a minute, which is denoted by (°).

Also, a minute is divided into 60 equal parts and each part is called a second, which is denoted by (“). Thus.

1 right angle = 90° (degrees)
1° = 60′ (minutes) and
1′ = 60” (seconds).

Class 10 Maths Solutions Wbbse

2. Centesimal system:

In this system of measurement of angles, 1 right angle is taken as unit. Here 1 right angle is divided into 100 equal parts and each part is called a grade, which is denoted by (g).

A grade is divided into 100 equal parts and each part is called a minute , which is denoted by (‘). A minute is further divided into 100 equal parts and each part is called a second, which is denoted by (“). Thus.

1 right angle = 100g (grades)
1g = 100′ (minutes)
1′ =100″ (seconds)

[1 means 1 sexagesimal minute, while 1′ means 1 centesimal minute. Similarly, l” means 1 sexagesimal second while 1″ means 1 centesimal second.]

Circular or radian system:

In this system 1 radian is taken as unit. So, you may ask what is a radian?

Definition of radian:

In any circle the angle subtended at the centre by an arc equal to the radius of the circle is called a radian and is denoted by (1c).

Thus 1c means an angle equal to 1 radian. The most important criterion of radian angles is that in all circles the ratio of the circumference to the diameter is always constant.

The value of this constant ratio is denoted by π, a Greek letter which is pronounced as Pi. π is an incommensurable number approximately equal to \(\frac{22}{7}\).

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles Theorems

Theorem 1: A radian is a constant angle.

Or,

The angle subtended at the centre of a circle by an arc which is equal in length to the radius of the circle is constant.

Let APC be a circle whose centre is O and let \(\widehat{A P}\) be an arc = radius OA = r.

Then ∠AOP = 1c (one radian).

We have to prove that ∠AOP is constant;

Class 10 Maths Solutions Wbbse

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Theorem A Radian Is A Constant Angle

 

Construction:

Let us draw OB perpendicular to OA, then the arc AB = \(\frac{1}{4}\) of the circumference.

Proof: The angles at the centre of a circle are proportional to the corresponding arcs subtended by

∴ \(\frac{\angle A O P}{\angle A O B}=\frac{{arc} A P}{{arc} A B}=\frac{\text { radius } \mathrm{OA}}{\frac{1}{4} \text { of the circumference }}\)

= \(\frac{r}{\frac{1}{4} \times 2 \pi r}=\frac{2}{\pi}\)

Hence \(\frac{\angle \mathrm{AOP}}{\angle \mathrm{AOB}}=\frac{2}{\pi}\)

or, \(\frac{\angle \mathrm{AOP}}{1 \text { right angle }}=\frac{2}{\pi}\) [∵ ∠AOB = 1 right angle]

or, ∠AOP= \(\frac{2}{\pi}\) right angle.

or, 1c = \(\frac{2}{\pi}\) right angle

But 2 and π are constants.

∴ 1c = constant [∵ right angles are always constant ]

Hence a radian is a constant angle. [ Proved ]

Theorem 2. The circular measure of an angle is equal to the ratio of the arc of any circle subtending that angle at its centre to the radius of- the circle.

or,

Prove that the radian measure of any angle at the centre of a circle is expressed by the fraction subtending arc radius.

Let MON be an angle, whose circular measure is to be determined.

 

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Theorem 2

 

Construction: Let us draw a circle with centre at O and with any radius r.

Let the circle intersects the sides OM and ON at the points A and B respectively.

Let AP be an arc equal to the radius r. Let us join O and P.

Proof: ∠AOP = 1 radian.

Since the angles at the centre of a circle are proportional to the arcs which subtend them,

∴ \(\frac{\angle \mathrm{AOB}}{\angle \mathrm{AOP}}=\frac{{arc} \mathrm{AB}}{{arc} \mathrm{AP}}=\frac{{arc} \mathrm{AB}}{r}\) [because arc AP =r]

Now, ∠MON =∠AOB = \(\frac{{arc} \mathrm{AB}}{r}\) x ∠AOP

= \(\frac{{arc} \mathrm{AB}}{r}\) x 1 radian [∵ ∠AOP = 1 radian] = \(\frac{{arc} \mathrm{AB}}{r}\) radian

Hence the theorem. [ Proved ]

Relation among the three systems:

1. Relation between sexagesimal and centesimal system:

1 right angle = 90° and 1 right angle = 100g

∴ 90° (degree) = 100g (grade)

or, \(1^{\circ}=\left(\frac{100}{90}\right)^g=\left(\frac{10}{9}\right)^g \text { and } 1^g=\left(\frac{90}{100}\right)^{\circ}=\left(\frac{9}{10}\right)^{\circ}\)

2. Relation between sexagesimal and radian system:

1 right angle = 90° and 1 right angle = \(\frac{\pi^c}{2}\)(radian)

∴ 90° = \(\frac{\pi^c}{2}\) or, \(\frac{\pi^c}{180}\)

and πc = 180° or, \(1^c=\left(\frac{180}{\pi}\right)^{\circ}\)

3. Relation between centesimal and radian system:

1 right angle = 100g and 1 right angle = \(\frac{\pi^c}{2}\)

∴ \(100^g=\frac{\pi^c}{2}\)

or, \(1^g=\frac{\pi^c}{200}\)

∴ \(\pi^c=200^g\)

or, \(1^c=\left(\frac{200}{\pi}\right)^g\)

From the above (1), (2) and (3) relations we can write, 180°= 200g = πc

Also, 1 radian = \(\frac{180^{\circ}}{\pi}=\frac{180^{\circ}}{\frac{22}{7}}\left[because \pi=\frac{22}{7}\right]\)

= \(\frac{180^{\circ} \times 7}{22}\) = 57° 16′ 22″ (approximately)

Again, \(1^{\circ}=\left(\frac{\pi}{180}\right)^c=\left(\frac{\frac{22}{7}}{180}\right)^c=\left(\frac{22}{7 \times 180}\right)^c<1^c\)

Hence, 1° < 1c i.e., 1g > 1°.

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles Multiple Choice Questions

Example 1. The end point of the minute hand of a clock rotates in 1 hour

  1. \(\frac{\pi}{4}\) radian
  2. \(\frac{\pi}{2}\) radian
  3. π radian
  4. 2π radian

Solution: 4. 2π radian.

We know that the end point of the minute hand of a clock rotates in. 1 horn a complete resolution of the clock i.e., it starts at the marking point 12 and ends at the same point.

So a complete revolution is performed.

It means that the minute, hand rotates 360° in 1 hour.

Now, 360° = \(\left(360^{\circ} \times \frac{\pi}{180^{\circ}}\right)\) radian = 2π radian.

Example 2. \(\frac{\pi}{6}\) radians equals to

  1. 60°
  2. 45°
  3. 90°
  4. 30°

Solution: 4. 30°

We know that πc = 180°

∴ \(1^c=\frac{180^{\circ}}{\pi}\)

∴ \(\frac{\pi^c}{6}=\frac{180^{\circ}}{\pi} \times \frac{\pi}{6}=30^{\circ}\)

Hence \(\frac{\pi}{6}\) radian = 30°

Example 3. The circular value of each internal angle of a regular hexagon is

  1. \(\frac{\pi}{3}\)
  2. \(\frac{2\pi}{3}\)
  3. \(\frac{\pi}{6}\)
  4. \(\frac{\pi}{4}\)

Solution: 2. \(\frac{2\pi}{3}\)

The number of sides of a, regular hexagon is 6.

∴ The value of each external angle = \(\frac{360^{\circ}}{6}\) = 60°

∴ The value of each internal angle = 180° – 60° = 120° =120° x \(\frac{\pi}{180^{\circ}}\) = \(\frac{2 \pi}{3}\)

Hence the required value is \(\frac{2 \pi}{3}\).

Example 4. The measurement of 0 in the relation s = rQ is determined by

  1. Sexagesimal system
  2. Circular system
  3. Those two systems
  4. None of these two systems

Solution: 2. Circular system.

Example 5. In cyclic quadrilateral ABCD, if ∠A = 120°, then the circular value of ∠C is

  1. \(\frac{\pi}{3}\)
  2. \(\frac{\pi}{6}\)
  3. \(\frac{\pi}{2}\)
  4. \(\frac{2\pi}{3}\)

Solution: 1. \(\frac{\pi}{3}\)

Here, ∠A and ∠C are opposite angles of the cyclic quadrilateral ABCD.

∴ ∠A + ∠C = 180°

or, 120° + ∠C = 180° or, ∠C = 180° – 120° = 60°

∴ ∠C = 60° x \(\frac{\pi}{180^{\circ}}\) = \(\frac{\pi}{3}\)

Hence the required circular value = \(\frac{\pi}{3}\)

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles True Or False

Example 1. The angle, formed by rotating a ray centering, its end point in anticlockwise direction is positive.

Solution: True

since by definition, the anti-clockwise rotative angle is positive.

Example 2. The angle, formed for completely rotating a ray twice by centering its end point is 720°.

Solution: True

since 1 complete rotation = 360°

∴ 2 complete rotation = 360° x 2 = 720°.

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles Fill In The Blanks

Example 1. π radian is a ______ angle.

Solution: Constant

Example 2. In sexagesimal system 1 radian equals to _______ (approximately).

Solution: 57°16’22”

Example 3. The circular value of the supplementary angle of the measure \(\frac{3\pi}{8}\) is ______

Solution: \(\frac{5\pi}{8}\)

since, \(\pi-\frac{3 \pi}{8}=\frac{8 \pi-3 \pi}{8}=\frac{5 \pi}{8}\)

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles Short Answer Type Questions

Example 1. If the value of an angle in degree is D and in radiap is R ; then determine the value of \(\frac{R}{D}\)

Solution:

Given:

If the value of an angle in degree is D and in radiap is R

We know that 1 radian = \(=\left(\frac{\pi}{180}\right)^0\)

∴ R radian = \(\frac{R \pi}{180^{\circ}}\)

∴ D = \(\frac{R \pi}{180}\)

Again, \(=\frac{\mathrm{R}}{\frac{\mathrm{R} \pi}{180}}=\frac{180}{\pi}\)

Hence the required value of \(\frac{\mathrm{R}}{\mathrm{D}}\) = \(\frac{180}{\pi}\).

Example 2. Determine the value of the complementary angle of the measure 63° 35′ 15″.

Solution: The complementary angle of 63° 35′ 15″ = 90° – 63° 35′ 15″ = 26° 24′ 45″.

Example 3. If the measures of two angles of a triangle are 65° 56′ 55″ and 64° 3′ 5″, then calculate the circular value of the third angle.

Solution:

Given:

If the measures of two angles of a triangle are 65° 56′ 55″ and 64° 3′ 5″,

Third angle = 180° – (65°56’55” + 64°3’5″) = 180° – 130° = 50°

= 50° x \(\frac{\pi}{180^{\circ}}=\frac{5 \pi}{18}\)

Hence the required circular measure = \(\frac{5 \pi}{18}\).

Example 4. In a circle, if an arc of length 220 cm subtends an angle of measure 63° at the centre, then determine the radius of the circle.

Solution:

Given:

In a circle, if an arc of length 220 cm subtends an angle of measure 63° at the centre

63° is made by the arc 220 cm

∴ 1° is made by the arc \(\frac{220}{63}\) cm

∴ 360° is made by the arc \(\frac{220 \times 360}{63}\) cm = \(\frac{8800}{7}\) cm

∴ the perimeter of the circle = \(\frac{8800}{7}\) cm

Let r be the radius of the circle.

∴ \(2 \pi r=\frac{8800}{7}\)

or, \(2 \times \frac{22}{7} \times r=\frac{8800}{7}\) or, r = 200.

Hence the radius of the circle = 200 cm.

Example 5. Find the circular value of an angle formed by the end point of hour hand of a clock in 1 hour rotation.

Solutions: In 1 hour rotation the hour hand of a clock rotates = \(\frac{360^{\circ}}{12}\) = 30° [∵ 12 hour rotates 360°]

= \(30^{\circ} \times \frac{\pi}{180^{\circ}}=\frac{\pi}{6}\)

Hence the required circular value = \(\frac{\pi}{6}\).

 

Trigonometry Chapter 1 Concept Of Measurement Of Angles Long Answer Type Questions

Example 1. In ΔABC, AC = BC, and BC is extended upto point D. If ∠ACD = 144°, then determine the circular value of each of the angles of ΔABC.

Solution:

Given:

In ΔABC, AC = BC, and BC is extended upto point D. If ∠ACD = 144°

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Long Answer Question Example 1

 

In ΔABC, AC = BC,

∴ ∠ABC = ∠BAC

∠ACD = 144°; ∠ACB = 180° – ∠ACD = 180° – 144° = 36°

Again, ∠ABC + ∠BAC + ∠ACB = 180°

or, ∠ABC + ∠ABC + 36° = 180° [∵ ∠BAC = ∠ABC]

or, 2∠ABC = 180° – 36°

or, 2∠ABC = 144°

or, ∠ABC = \(\frac{144^{\circ}}{2}\) = 72°

∴ ∠BAC = ∠ABC = 72°.

Hence three angles of the triangle ΔABC are 72°, 72°, 36°

or, \(72^{\circ} \times \frac{\pi}{180^{\circ}}, 72^{\circ} \times \frac{\pi}{180^{\circ}}, 36^{\circ} \times \frac{\pi}{180^{\circ}}\)

or, \(\frac{2 \pi}{5}, \frac{2 \pi}{5}, \frac{\pi}{5}\)

The circular value of each of the angles of ΔABC = \(\frac{2 \pi}{5}, \frac{2 \pi}{5}, \frac{\pi}{5}\)

Example 2. If the difference of two acute angles of a right angled triangle is \(\frac{2\pi}{5}\), then find the sexagesimal values of two angles.

Solutions:

Given:

If the difference of two acute angles of a right angled triangle is \(\frac{2\pi}{5}\)

Let the two angles be x° and y°.

∴ x – y = \(\frac{2\pi}{5}\) or, x – y = \(\frac{2 \times 180}{5}\)

or, x-y =72…….(1)

Also, x + y = 90……..(2)

Adding (1) and (2) we get, 2x = 162 or, x = \(\frac{162}{2}\) = 81

∴ 81 + y = 90 or, y = 90 – 81 = 9

Hence the required angles are 81° and 9°.

Example 3. The measure of one angle of a triangle is 65° and other angle is \(\frac{\pi}{15}\), then determone the sexagesimal value and circular value of third angle.

Solution:

Given:

The measure of one angle of a triangle is 65° and other angle is \(\frac{\pi}{15}\),

\(\frac{\pi}{12}=\frac{180^{\circ}}{12}\) = 15°

∴ the third angle = 180° – (65° + 15°) = 180° – 80°= 100°

= 100° x \(\frac{\pi}{180^{\circ}}\) = \(\frac{5\pi}{9}\)

Hence the third angle is 100° or \(\frac{5\pi}{9}\).

Example 4. If the sum of two angles is 135° and their difference is then determine the sexagesimal value and circular value of two angles.

Solution:

Given:

The sum of two angles is 135°.

Let the two angles be x° and y° (x > y).

As per question, x + y = 135………(1)

and x-y = \(\frac{\pi}{12}\) = \(\frac{180}{12}\) = 15…….(2)

Now, adding (1) and (2) we get, 2x = 150

or, x = \(\frac{150}{12}\) = 75 = 75 x \(\frac{\pi}{100}\) = \(\frac{5\pi}{12}\)

From(1) we get, 75 + y = 135 -75 or, y = 60 = 60 x \(\frac{\pi}{180}\) = \(\frac{\pi}{3}\)

Hence the sexagesimal values of the angles are 75° and 60° and the circular values are \(\frac{5\pi}{12}\) and \(\frac{\pi}{3}\).

Example 5. If the ratio of three angles of a triangle is 2 : 3: 4, then determine the circular value of the greatest angle.

Solution:

Given:

The ratio of three angles of a triangle is 2 : 3: 4.

Let the angles be 2x°, 3x° and 4x°.

∴ 2x° + 3x° + 4x° = 180°

or, 9x° = 180° or, x° = \(\frac{180^{\circ}}{9}\) = 20°

∴ the greatest angle = 4x° = 4 x 20° = 80°

= 80° x \(\frac{\pi}{180^{\circ}}=\frac{4 \pi}{9}\)

Hence the circular value of the greatest angle is \(\frac{4 \pi}{9}\).

Example 6. The length of a radius of a circle is 28 cm. Determine the circular value of angle subtended by an arc of 5*5 cm length at the centre of this circle.

Solution:

Given:

The length of a radius of a circle is 28 cm.

The cicumfercnce of the circle = 2 x \(\frac{22}{7}\) x 28 cm =176 cm

∴ 176 cm subtends 360° at the centre

∴ 1 cm subtends \(\frac{360^{\circ}}{176}\) at the centre

∴ 5.5 cm subtends \(\frac{360^{\circ} \times 5 \cdot 5}{176}\) at the centre

= \(\frac{180^{\circ}}{16} \times \frac{\pi}{180^{\circ}}\)

= \(\frac{\pi}{16}\) at the centre

Hence the required circular value \(\frac{\pi}{16}\).

Example 7. The ratio of two angles subtended by two arcs of unequal lengths at the centre is 5: 2 and if the sexagesimal value of the second angle is 30°, then detrmine the sexagesimal value and the circular value of the first angle.

Solution:

Given:

The ratio of two angles subtended by two arcs of unequal lengths at the centre is 5: 2 and if the sexagesimal value of the second angle is 30°

Let the first angle be x°.

As per question, x° : 30° = 5:2

or, \(\frac{x}{30}=\frac{5}{2}\)

or, 2 x=150 or, x = \(\frac{150}{2}=75\)

∴ the sexagesimal value of the first angle is 75° and the circular value is

75° x \(\frac{\pi}{180^{\circ}}=\frac{5 \pi}{12}\).

Example 8. A rotating ray makes an angle -5 \(\frac{1}{2}\)1t. Determine the direction in which the ray has completely rotated and there after what more angle it has produced.

Solution:

Given:

A rotating ray makes an angle -5 \(\frac{1}{2}\)1t.

\(-5 \frac{1}{12} \pi=-\frac{61}{12} \times 180^{\circ}\) = -61 x 15° = -915°

We know that 1 complete rotation = 360°

∴ 915° = 360° x 2 + 195°

∴ 2 complete rotation clockwise and thereafter makes 195° more.

Example 9. Reme has drawn an isosceles triangle ABC whose included angle of two equal sides is ∠ABC = 45° the bisector of ∠ABC intersects the side AC at the point D. Determine the circular values of ∠ABD, ∠BAD, ∠CBD, and ∠BCD.

Solution:

Given:

Reme has drawn an isosceles triangle ABC whose included angle of two equal sides is ∠ABC = 45° the bisector of ∠ABC intersects the side AC at the point D.

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Long Answer Question Example 9

 

Let BD is the bisector of ∠ABC which intersects the sides AC at D.

∴ ∠ABD = \(\frac{1}{2}\)∠ABC = \(\frac{1}{2}\) x 45° [∵ ∠ABC = 45°]

= \(\frac{1}{2}\) x 45° x \(\frac{\pi}{180^{\circ}}\) = \(\frac{\pi}{8}\)

∠BAD = 90° – ∠ABD

= 90° – \(\frac{45^{\circ}}{2}=\frac{180^{\circ}-45^{\circ}}{2}\)

= \(\frac{135^{\circ}}{2} \times \frac{\pi}{180^{\circ}}=\frac{3 \pi}{8}\)

[∵ BD is the bisector of ∠ABC and AB = BC, ∴ BD ⊥ AC and ∠ADB = 90°]

∠CBD = ∠ABD = \(\frac{\pi}{8}\)

∠BCD = ∠BAD [∵ BA = BC, ∠BCD – ∠BAD] = \(\frac{3\pi}{8}\)

Hence the required angles are \(\frac{\pi}{8}\), \(\frac{3\pi}{8}\), \(\frac{\pi}{8}\), \(\frac{3\pi}{8}\) respectively.

Example 10. The base BC of the equilateral triangle ABC is extended upto the point E so that CE = BC. By joining A, E, determine the circular values of the angles of ΔAEC.

Solutions:

Given:

The base BC of the equilateral triangle ABC is extended upto the point E so that CE = BC. By joining A, E,

WBBSE Solutions For Class 10 Maths Trigonometry Chapter 1 Concept Of Measurement Of Angles Long Answer Question Example 10

 

Since ΔABC is equilateral,

∴ ∠ACB = 60°.

∴ ∠ACE = 180° – ∠ACB

= 180° – 60° = 120°

= 120° x \(\frac{\pi}{180^{\circ}}=\frac{2 \pi}{3}\)

∵ CE = BC (given) = AC [∵ ΔABC is equilateral ]

∴ ∠CAE = ∠AEC……..(1)

Again, ∠ACE + ∠CAE + ∠AEC = 180°

or, 120° + ∠AEC + ∠AEC = 180° [ by (1) ]

or, 120° + 2∠AEC = 180° or, 2 ∠AEC = 180° – 120°

or, 2∠AEC = 60°

or, ∠AEC = \(\frac{60^{\circ}}{2}\) = 30° = 30° x \(\frac{\pi}{180^{\circ}}=\frac{\pi}{6}\)

∴ ∠AEC = \(\frac{2\pi}{3}\), ∠AEC = \(\frac{\pi}{6}\) and ∠CAE = ∠AEC = \(\frac{\pi}{6}\) [by (1)]

Hence the required angles are \(\frac{2\pi}{3}\), \(\frac{\pi}{6}\) and \(\frac{\pi}{6}\)

Example 11. If the measures of three angles of a quadrilateral are \(\frac{\pi}{3}\), \(\frac{5\pi}{6}\) and 90° respectively, determine the sexagesimal and circular value of fourth angle.

Solutions:

Given:

If the measures of three angles of a quadrilateral are \(\frac{\pi}{3}\), \(\frac{5\pi}{6}\) and 90° respectively,

We have, 90° = 90° x \(\frac{\pi}{180^{\circ}}\) = \(\frac{\pi}{180}\)

Also, we know that the sum of the four angles of a quadrilateral is 360° = 2 x 180° = 2π.

∴ the fourth angle = \(2 \pi-\left(\frac{\pi}{3}+\frac{5 \pi}{6}+90^{\circ}\right)\)= \(2 \pi-\left(\frac{\pi}{3}+\frac{5 \pi}{6}+\frac{\pi}{2}\right)\)

= \(2 \pi-\frac{2 \pi+5 \pi+3 \pi}{6}=2 \pi-\frac{5 \pi}{3}\)

= \(\frac{6 \pi-5 \pi}{3}=\frac{\pi}{3}\)

Also, \(\frac{\pi}{3}\) = \(\frac{180^{\circ}}{3}\) = 60°.

Hence the sexagesimal and circular value of the fourth angle are 60° and y respectively.

Example 12. The angles of a triangle are successively in equal differnce. If the number of degrees in the greatest angle be same as the number of grades in the least one. Find the angles in degrees.

Solutions:

Given:

The angles of a triangle are successively in equal differnce. If the number of degrees in the greatest angle be same as the number of grades in the least one.

Let the angles are (a – d), a, (a + d) degrees respectively.

∴ a-d+a+a+d= 180° [∵ the sum of angles of a triangle is 180°]

or, 3a = 180° or, a = \(\frac{180^{\circ}}{3}\) = 60°

Here, the least angle is (a – d) degree

= \(\frac{10}{9}\)(a-d) grades [∵ 90° = 100g ] and the greatest angle is (a + d) degrees.

As per question, a + d = \(\frac{10}{9}\) (a- d)

or, 9a + 9d = 10a – 10d

or, 9d + l0d = 10a – 9a

or, 19d = a

or, 19d = 60 [∵ a = 60 ]

or, d = \(\frac{60}{19}\)

Hence the angles are degrees, (60 – \(\frac{60}{19}\)) degrees and (60 + \(\frac{60}{19}\)) degrees,

or, 56\(\frac{16}{19}\) degrees, 60 degrees and 63\(\frac{3}{19}\) degrees.

Example 13. Find the times between 4 o’clock and 5 o’clock when the angle between the minute-hand and hour-hand is \(\frac{8\pi}{15}\) radians.

Solutions: \(\frac{8\pi}{15}\) radians = \(\frac{8 \times 180^{\circ}}{15}\) = 96°

We know that the circumference of clock is equal to 60-minute divisions.

∴ 360° angle is subtended by 60-minute divisions

∴ 1° angle is subtended by \(\frac{60}{360^{\circ}}\) minute divisions

∴ 96° angle is subtended by \(\frac{60 \times 96^{\circ}}{360^{\circ}}\) minute divisions

= 16-minute divisions

∴ The distance between two hands will be 16-minute divisions.

Now, at 4 o’clock the minute, hand was 20-minute divisions behind the hour-hand.

So, the minute hand gains either (20 – 16) = 4-minute divisions or (20 + 16) = 36-minute divisions.

We know that the minute-hand gains 55-minute divisions in 60 minutes

∴ the minute-hand gains 4 minute divisions in \(\frac{60 \times 4}{55}\) minutes = \(\frac{48}{11}\) minutes = 4 \(\frac{4}{11}\) minutes

Also, the minute hand gains 36-minute divisions in \(\frac{60 \times 36}{55}\) minutes = 39\(\frac{3}{11}\)minutes

Hence the required time will be 4\(\frac{4}{11}\)minutes past 4 o’clock and 39\(\frac{3}{11}\)minutes past 4 o’clock.

 

 

 

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects

Mensuration Chapter 5 Problems Related To Different Solids And Objects

Rectangular parallelopiped

In the length, breadth and height ofa rectangular parallelopiped be a unit, b unit and c unit respectively,then

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Rectangular Parallelopiped

 

Total surface area of the rectangular parallelopiped

= 2 x [Length x Breadth + Breadth x Height + Height x Length] sq-unit

Total surface area of the rectangular parallelepiped = 2 x (ab + be + ca) sq-unit.

Length of the diagonal of the rectangular

parallelopiped = \(\sqrt{(\text { Length })^2+(\text { Breadth })^2+(\text { Height })^2}\)unit

= \(\sqrt{a^2+b^2+c^2}\)unit

WBBSE Solutions for Class 10 Maths

Volume of the rectangular parallelopiped

= (Area of the base) x Height cubic-unit

= Length x Breadth x Height cubic unit

Volume of the rectangular parallelepiped = abc cubic-unit.

Cube:

Maths Solutions Class 10 Wbbse

If each side of a cube be a unit, then

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Cube

 

The total surface area of the cube

= 6 x (side)2 sq-unit

The total surface area of the cube= 6a2 sq-unit

Length of the diagonal of the cube

= √3 x (Length of side) unit

Length of the diagonal of the cube = √3 a unit

Volume of the cube = (side)3 cubic-unit

Volume of the cube = a3 cubic-unit.

Right circular cylinder

Wbbse Class 10 Maths Solutions

If the radius of the base of a right circular cylinder be r unit and its height be h unit, then

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Right Circular Cylinder

 

Curved surface area of the cylinder

= (circumference of the base) x height sq-unit

= 2πr x h sq-unit

Curved surface area of the cylinder = 2πrh sq-unit

Total surface area of the cylinder

= (curved surface area) + (area of two end-circle) sq-unit

= (2πrh + 2πr2) sq-unit

Total surface area of the cylinder = 27π(h + r) sq-unit

Volume of the cylinder

Maths Solutions Class 10 Wbbse

= (Area of the circular base) x height cubic-unit

= πr2 x h cubic-unit

Volume of the cylinder = πr2h cubic-unit.

Right circular cone:

If the radius of the base be r unit, height be h unit and slant height be l unit, then

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Right Circular Cone

 

Curved surface area of the cone

= \(\frac{1}{2}\) x (circumference of the base) x slant height sq-unit

= \(\frac{1}{2}\) x 2πr x l sq-unit

Curved surface area of the cone = πrl sq-unit.

Total surface area of the cone

= (curved surface area) + (area of the base) sq-unit

= (πrl +πr2) sq-unit

Total surface area of the cone = πr (l + r) sq-unit.

Volume of the cone

= \(\frac{1}{3}\) x (area of the base) x height cubic-unit

= \(\frac{1}{3}\) x (πr2) x h cubic-unit

Volume of the cone = \(\frac{1}{3}\) πr2h cubic-unit

Maths Solutions Class 10 Wbbse

Slant height of the cone

= \(\sqrt{(\text { radius of base })^2+(\text { height })^2}\) unit

Slant height of the cone = \(\sqrt{r^2+h^2}\) unit.

Sphere:

If the radius of the sphere be r unit, then

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Sphere

 

Total surface area of the sphere

= π x (diameter)2 sq-unit

= π x (2r)2 sq-unit

= π x 4r2 sq-unit

Total surface area of the sphere = 4πr2 sq-unit

Volume of the sphere

= \(\frac{4}{3}\) x π x (Radius)3 cubic-unit

Volume of the sphere = \(\frac{4}{3}\)πr3 cubic-unit.

Hemisphere

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Hemisphere

 

If the radius of a hemisphere, be r unit, then

Total surface area of a hollow hemisphere = 2πr2 sq-unit

Curved surface area of a solid hemisphere = 2πr2 sq-unit

Total surface area of a solid hemisphere = 3πr2 sq-unit

Volume of a solid or hollow hemisphere = \(\frac{2}{3}\)πr3 cubic-unit

 

Mensuration Chapter 5 Problems Related To Different Solids And Objects Multiple Choice Questions

Example 1. If a solid right circular cone of height r unit is made by melting a solid sphere of radius r unit, then the radius of the base of the cone will be

  1. 2r unit
  2. 3r unit
  3. r unit
  4. 4r unit

Solution: Radius of the sphere = r unit

∴ Volume of the sphere = \(\frac{4}{3}\) πr3 cubic-unit.

Also, the height of the cone = r unit

Let the radius of the base of the cone be r1 unit

∴ Let the radius of the base = r1 unit

∴ the volume of the cone = \(=\frac{1}{3} \pi r_1^2\) x r cubic-unit

As per question, \(\frac{1}{3} \pi r_1^2 \times r=\frac{4}{3} \pi r^3\)

⇒ \(r_1^2=4 r^2\)

⇒ \(r_1^2=(2 r)^2\)

⇒ r1 = 2r

∴ radius of the cone = 2r unit

Hence 1. 2r unit is correct

Example 2. The radius of a right circular cylinder is r un it and height is 2r units. The diameter of the largest sphere that can be put into the cylinder will be

Wbbse Class 10 Maths Solutions

  1. r unit
  2. 2r unit
  3. \(\frac{r}{2}\)
  4. 4r unit

Solution: The radius of the cylinder = r unit and height = 2r unit.

∴ the diameter of the sphere = 2r unit.

Hence 2. 2r unit is correct.

Example 3. The volume of the largest solid cone that can be cut off from a solid hemisphere will be

  1. 4πr3 cubic-unit
  2. 3πr3 cubic-unit
  3. \(\frac{\pi r^3}{4}\) cubic-unit
  4. \(\frac{\pi r^3}{3}\) cubic-unit

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Solid Cone Can Be Cut Off From Solid Hemisphere

 

Solution: The radius of the hemisphere = r unit

∴ the radius of the cone = r unit

and height of the cone = r unit

∴ volume of the cone = \(\frac{1}{3}\)πr2 x r cubic-unit

= \(\frac{\pi r^3}{3}\) cubic unit

Hence 4. \(\frac{\pi r^3}{3}\) cubic-unit is correct.

Example 4. The radius of the largest sphere that can be cut off from a solid cube of side x unit each will be

  1. x unit
  2. 2x unit
  3. \(\frac{x}{2}\) unit
  4. 4x unit

Solution: The length of each side of the cube = x unit

∴ the diameter of the largest sphere = x unit

∴ the radius of the largest sphere = \(\frac{x}{2}\) unit

Hence 3. \(\frac{x}{2}\) unit is correct.

Example 5. The height of the water level when the water of a right circular cone-shaped bottle of radius r unit and of height h unit is poured into a right circular cylinder-shaped pot of radius mr unit will be

  1. \(\frac{h}{2} m^2\) unit
  2. \(\frac{2 h}{m}\) unit
  3. \(\frac{h}{3 m^2}\) unit
  4. \(\frac{m}{2 h}\) unit

Solution: The radius of the conical bottle = r unit and height = h unit

∴ the volume of the bottle = \(\frac{1}{3}\) πr2h cubic-unit.

Again, the radius of the cylindrical pot = mr unit,

Let the water-level will rise x unit.

∴ the volume of the water poured into the pot = π x (mr)2 x x cubic-unit

∴ π(mr)2x = \(\frac{1}{3}\)h

⇒ m2r2 x = \(\frac{r^2 h}{3}\)

⇒ x = \(\frac{h}{3 m^2}\)

the water-level of the pot will rise unit.

Hence 3. \(\frac{h}{3 m^2}\) unit is correct.

Wbbse Class 10 Maths Solutions

Mensuration Chapter 5 Problems Related To Different Solids And Objects True Or False

Example 1. If two solid hemispheres of same type of which the radii of bases are r unit be joined along their bases, then the total surface area of the joint solid object is 6πr2 square-unit.

Solution: False

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Two Solid Hemispheres Are Same Type

 

Since the radii of the hemispheres is r unit and the two hemispheres are joint along their bases, so we shall get a full sphere, the total surface area which will be 4πr2sq-unit

Example 2. The radius of base, height and slant height of a solid right circular cone are r unit, h unit and l unit respectively. The base of the cone is joint along the base of solid right circular cylinder. If the radii of the bases and heights of the cylinder and of the cone be equal, then the total surface area of the joint solid object will be (πrl + 2πrh + 2πr2) sq-unit.

Solution: False.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Radius Of Base height And Slant Height

 

Since the radius of the base of the cone is r unit and its slant height is l unit.

∴ curved surface area = πrl sq-unit.

Again, the radius of the cylinder is r unit and height is h unit.

∴ the surface area of the solid object = (πrl + 2πrh + πr2) sq-unit.

Mensuration Chapter 5 Problems Related To Different Solids And Objects Fill In The Blanks

Example 1. The radii of bases of a solid right circular cone and of two hemispheres are equal. If the two hemispheres be joined with two end-planes of the cone, then the total surface area of the new solid object = (the curved surface area of one of the hemispheres) + (curved surface area of the ______ ) + (the curved surface area of the other hemisphere).

Solution: cylinder

Example 2. The shape of a pencil with its one end cut is the coordination of a right circular cone and a right circular ______

Solution: cylinder

Example 3. If a solid sphere is melted to make a solid right circular cylinder, then the volumes of the sphere and the cylinder are ______

Solution: equal

Mensuration Chapter 5 Problems Related To Different Solids And Objects Short Answer Type Questions

Example 1. If a solid sphere of radius r unit be melted to make a solid right circular cone of height r unit, then find the radius of the base of the cone.

Solution:

Given:

A solid sphere of radius r unit be melted to make a solid right circular cone of height r unit

The radius of the sphere = r unit.

∴ the volume of the sphere = \(\frac{4}{3}\)πr3 cubic-unit.

Again, the height of the cone = r unit

Let the radius of the cone = x unit.

∴ the volume of the cone = \(\frac{1}{3}\)πx2r cubic-unit.

As per question, \(\frac{1}{3}\)πx2r = \(\frac{4}{3}\)πr3

⇒ x2 = 4r2

⇒ x2 = (2r)2  ⇒ x = 2r

Hence the radius of base of the cone is 2r unit.

Wbbse Class 10 Maths Solutions

Example 2. If the radii and volumes of a solid right circular cone and a solid sphere be equal, then what will be the ratio of the length of diameter of the sphere and the height of the cone?

Solution:

Given:

The radii and volumes of a solid right circular cone and a solid sphere be equal

Let the radii of the cone and the sphere be r unit and the height of the cone be x-unit.

∴ the volume of the cone = \(\frac{1}{3}\)πr2x cubic-uinit

and the volume of the sphere = \(\frac{4}{3}\)πr3 cubic-unit.

As per question,\(\frac{4}{3}\)πr3 = \(\frac{1}{3}\)πr2x

⇒ 4r = x

⇒ 2r = \(\frac{x}{2}\) = \(\frac{1}{2}\)

⇒ 2r: x = 1 : 2

Hence. the required ratio is 1:2.

Example 3. The shape of the lower part of a solid object is like a hemisphere and the upper part like a right circular cone. If the surface area of the two parts be equal, then find the ratio of the length of radius and height of the cone.

Solution:

Given:

The shape of the lower part of a solid object is like a hemisphere and the upper part like a right circular cone. If the surface area of the two parts be equal

Let the radius of the hemisphere = r unit

∴ the radius of the cone = r unit

Let the height of the cone = h unit.

As per the question,

27πr2 = πrl, where l = slant height of the cone.

⇒ 2r = l ⇒ 2r = \(\sqrt{h^2+r^2}\)

⇒ 4r2 = h2 + r2  ⇒ 3r2 = h2

⇒ √3r = h

⇒ \(\frac{r}{h}=\frac{1}{\sqrt{3}}\)

⇒ r: h = 1: √3

Hence the required ratio = 1 : √3.

Example 4. The length ofradius of base of a solid right circular cone is equal to the length of radius of a solid sphere. If the volume of the sphere is 2 times the volume of the cone, then find the ratio of the height and radius of base of the cone.

Solution:

Given:

The length ofradius of base of a solid right circular cone is equal to the length of radius of a solid sphere. If the volume of the sphere is 2 times the volume of the cone

Let the radius of the base of the cone = r unit

∴ the radius of the sphere = r unit

Let the height of the cone = h unit

∴ the volume of the cone = \(\frac{1}{3}\)πr2h cubic-uinit

and the volume of the sphere = \(\frac{4}{3}\)πr3 cubic-unit.

As per question, 2 x \(\frac{1}{3}\)πr2h = \(\frac{4}{3}\)πr3

⇒ \(\frac{h}{r}=\frac{4}{2} \Rightarrow \frac{h}{r}=\frac{2}{1}\)

⇒h: r = 2: 1

Hence the required ratio is 2: 1.

Wbbse Class 10 Maths Solutions

Example 5. If a right circular cone of height 15 cm and of diameter 30 cm is made from a wooden sphere of radius 15 cm, then what percent of wood will be lost?

Solution:

Given:

A right circular cone of height 15 cm and of diameter 30 cm is made from a wooden sphere of radius 15 cm

The radius of the sphere = 15 cm

∴ the volume of the sphere = \(\frac{4}{3}\)π x (15)3 cc.

Now, the radius of the cone = \(\frac{30}{2}\) cm= 15 cm [∵ diameter = 30 cm]

and the height of the cone =15

∴ the volume of the cone = \(\frac{1}{3}\)π x (15)2 x 15 cc

= \(\frac{1}{3}\)π x (15)3 cc

∴ the lost wood = \(\left\{\frac{4}{3} \pi \cdot \times(15)^3-\frac{1}{3} \pi \times(15)^3\right\}\) cc

= \(\left(\frac{4}{3}-\frac{1}{3}\right) \pi \times(15)^3 \mathrm{cc}=\pi \times(15)^3 \mathrm{cc}\)

∴ the percent of lost wood = \(\frac{\pi \times(15)^3}{\frac{4}{3} \pi \times(15)^3}\) x 100% = 75%

Hence the required percentage = 75%.

Example 6. How many part of a rectangular parallelopiped type hole of length, breadth and depth 48 cm, 16.5 cm and 4 cm respectively should be stuffed with the soil obtained by digging a right circular conical tunnel of diameter 4 metres and of length 56 metres?

Solution: The volume of the soil obtained = π x \(\left(\frac{4}{2}\right)^2\) x 56 cubic-metre.

Again, the volume of the hole = 48 x 16.5 x 4 cubic-metre.

Let x part of the hole should be stuffed.

∴ x x 48 x 16.5 x 4 = π x 22 x 56

or, \(x=\frac{\frac{22}{7} \times 4 \times 56}{48 \times 16.5 \times 4}\)

or, x= \(\frac{22 \times 4 \times 8 \times 10}{48 \times 165 \times 4}\)

or, x = \(\frac{2}{9}\)

Hence \(\frac{2}{9}\) part of the hole should be stuffed.

Example 7. What will be the ratio of the diameter of the sphere and the height of a cylinder when the volumes of the sphere and the right circular cylinder are equal, given that the radii of the sphere and the cone are equal?

Solution: Let the radii of the sphere and the cylinder be r unit and the height of the cylinder be h unit.

As per question, πr2h = \(\frac{4}{3}\)πr3

⇒ 3h = 4r ⇒ 2.2r = 3h

⇒ \(\frac{2 r}{h}=\frac{3}{2}\)

⇒ 2r: h =3: 2

Hence the,ratio of the diameter of the sphere and the height of the cylinder is 3: 2.

Example 8. The ratio of the radii of a right circular cylinder and a right circular cone is 3: 4 and the ratio of their heights is 2 : 3..Then find the ratio of the volumes of them.

Solution:

Given:

The ratio of the radii of a right circular cylinder and a right circular cone is 3: 4 and the ratio of their heights is 2 : 3..

Let the radii Of the cylinder and the cone be 3r unit and 4r unit respectively and their heights be 2h unit and 3h unit respectively.

∴ Volume of the cylinder = π x (3r)2 x 2h cubic-unit = 18πr2h cubic-unit

and volume of the cone = \(\frac{1}{3}\)π x (4r)2 x 3h cubic-unit = 16πr2h cubic-unit

Hence the ratio of the volumes of cylinder and the cone is 18πr2h: 16πr2h

= \(\frac{18 \pi r^2 h}{16 \pi r^2 h}=\frac{9}{8}\) = 9: 8

Hence the required ratio = 9:8.

Example 9. How many spherical bullets of diameter 4 cm should be got from a cube of edges 44 cm?

Solution: The volume of the cube = (44)3 cc

The diameter of each spherical bullet = 4 cm

∴ the radius of each spherical bullet = \(\frac{4}{2}\)cm = 2 cm

the volume of each spherical bullet = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 23 cc

Let the number, of bullets to be got = x.

So, x = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 23 = (44)3

⇒ x = \(\frac{44 \times 44 \times 44 \times 3 \times 7}{4 \times 22 \times 2 \times 2 \times 2}\) = 2541

Hence the number of required bullets = 2541.

Example 10. If the curved surface areas of a sphere and a right circular cylinder both of same radii be equal, then what will be the ratio of their volumes?

Solution:

Given:

The curved surface areas of a sphere and a right circular cylinder both of same radii be equal

Let the radii of both the sphere and the cylinder be r unit and also let the height of the cylinder be h unit.

∴ the curved surface area of the sphere = 4πr2 sq-unit

and the curved surface area of the cylinder = 2πrh sq-unit.

As per the question, 4πr2 = 2πrh

or, \(\frac{\pi r^2}{\pi r h}=\frac{2}{4} \quad \text { or, } \quad \frac{r}{h}=\frac{1}{2}\)

or, h = 2r

∴ the ratio of the volumes of the sphere and the cylinder

= \(\frac{4}{3} \pi r^3: \pi r^2 h\)

= \(\frac{\frac{4}{3} \pi r^3}{\pi r^2 h}=\frac{4 \pi r^3}{3 \pi r^2 \times 2 r} \cdot\) [∵ because h=2 r]

= \(\frac{4 \pi r^3}{6 \pi r^3}=\frac{4}{6}=\frac{2}{3}\) =2: 3

Hence the required ratio of the volumes of sphere and the cylinder = 2:3.

Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Type Questions

Example 1. There is a solid iron-pillar in front of Anil’s house, the lower part of which is of type of a right circular cylinder and the upper part is of type of a right circular cone. The radii of their diameter of bases is 20 cm. If the height of the cylindrical part be 2.8 metres and that of the conical part be 42 cm and if the weight of1 cc iron be 7.5 gm, then what will be the weight of the pillar?

Solution:

Given:

There is a solid iron-pillar in front of Anil’s house, the lower part of which is of type of a right circular cylinder and the upper part is of type of a right circular cone. The radii of their diameter of bases is 20 cm. If the height of the cylindrical part be 2.8 metres and that of the conical part be 42 cm and if the weight of1 cc iron be 7.5 gm

The diameter of the bases = 20 cm

∴ The radii of the bases = \(\frac{20}{2}\) cm = 10 cm

The height of the cylindrical part = 2.8 metres = 280 cm.

∴ the volume of this part= \(\frac{22}{7}\) x (10)2 x 280 cc

= \(\frac{22}{7}\) x 100 x 280 cc

= 88000 cc

Again, the height of the conical part = 42 cm

∴ the volume of the conical part = \(\frac{1}{3}\) x \(\frac{22}{7}\) x (10)2 x 42 cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 100 x 42 cc

= 4400 cc

So, the total volume of the pillar = (88000 + 4400) cc = 92400 cc

∴ the weight of the total pillar = (7.5 x 92400) gm = 693000 gm = 693 kg

Hence the required weight = 693 kg.

Example 2. There is some water in a right circular cylindrical pot of diameter 24 cm. How much the height of water-level will be increased if 60 solid conical iron piece of diameter 6 cm and of height 4 cm each are completely immersed into that water?

Solution:

Given:

There is some water in a right circular cylindrical pot of diameter 24 cm.

The diameter of the conical iron-piece = 6 cm

∴ the radius of the conical iron piece = \(\frac{6}{2}\) cm = 3 cm

Also, the heights of the iron-pieces = 4 cm

So, the volume of each iron-piece = \(\frac{1}{2}\) x π x (3)2 x 4 cc = 12π cc

Again, the diameter of the conical pot = 24 cm

the radius of the conical pot = \(\frac{24}{2}\) cm = 12 cm

Let water-level will rise h cm

∴ the volume of the raised water= π x (12)2 x h cc = 144 πh cc

As per the question, 144 πh = 60 x 12π

⇒ h = \(\frac{60 \times 12 \pi}{144 \pi}\)

⇒ h = 5

Hence the water-level of the pot will rise 5 cm.

Example 3. The ratio of the curved surface areas of a solid right circular cone and of a right circular cylinder both of same radius of base and of same height is 5: 8. Then find the ratio of their radius of base and height.

Solution:

Given:

The ratio of the curved surface areas of a solid right circular cone and of a right circular cylinder both of same radius of base and of same height is 5: 8.

Let radius of base = r unit

and height of them = h unit

Also, let the slant height of the cone = l unit.

∴ curved surface area of the cone = πrl sq-unit

= \(\pi r \sqrt{r^2+h^2}\) sq-unit

The curved surface area of the cylinder = 2πrh sq-unit.

As per question, \(\frac{\pi r \sqrt{r^2+h^2}}{2 \pi r h}=\frac{5}{8}\)

or, \(\frac{\sqrt{r^2+h^2}}{h}=\frac{5}{4}\)

or, \(\frac{r^2+h^2}{h^2}=\frac{25}{16}\)

or, \(\frac{r^2}{h^2}+1=\frac{25}{16}\)

or, \(\frac{r^2}{h^2}=\frac{25}{16}-1\)

or, \(\frac{r^2}{h^2}=\frac{9}{16}\)

or, \(\frac{r}{h}=\sqrt{\frac{9}{16}}=\frac{3}{4}\)

⇒ r: h = 3: 4

Hence the required ratio of the radius of base and the height = 3: 4

Example 4. The radius of cross-section of a right circular rod is 3.2 dcm. The rod is nielted to make 21 solid spheres. If the radius of each sphere be 8 cm, then what will be the length of the rod?

Solution:

Given:

The radius of cross-section of a right circular rod is 3.2 dcm. The rod is nielted to make 21 solid spheres. If the radius of each sphere be 8 cm

The radius of the right circular rod = 3.2 dcm = 32 cm.

Let the length of the rod =x cm

∴ the volume of the rod = π x (32)2 x x cc

The radius of each spheres = 8 cm

∴ the volume of each sphere = \(\frac{4}{3}\) x π x (8)3 cc

So, the volume of 21 spheres = 21 x \(\frac{4}{3}\) x π x (8)3 cc

= 28 x 512 π cc

As per question, π x 32 x 32 x = 28 x 512 π

⇒ x = \(\frac{28 \times 512}{32 \times 32}\) = 14

Hence the required length of the rod = 14 cm.

Example 5. How many solid spheres of diameter 2.1 dcm each can be made by melting a solid rectangular parallelopiped copper piece of length 6.6 dcm, of breadth 4.2 dcm and of thickness 1.4 dcm? How many cubic-dcm metal will possess in each- sphere?

Solution: The length of copper piece = 6.6 dcm = 66 cm

The breadth of copper piece = 4.2 dcm = 42 cm

Height of copper piece= 1.4 dcm = 14 cm

So, the volume of coper piece = (66 x 42 x 14) cc

The diameter of each sphere = 2.1 dcm = 21 cm

∴ the radius of each sphere = \(\frac{21}{2}\) cm

∴ the volume of each sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x \(\left(\frac{21}{2}\right)^3\) cc

= \(\frac{4}{3} \times \frac{22}{7} \times \frac{21 \times 21 \times 21}{2 \times 2 \times 2}\)

= 11 x 21 x 21 cc

Let the number of spheres be x.

∴ x x 11 x 21 x 21 = 66 x 42 x 14

⇒ x = \(\frac{66 \times 42 \times 14}{11 \times 21 \times 21}\)

⇒ x = 8

Also, the volume of each sphere = 11 x 21 x 21cc

= \(\frac{11 \times 21 \times 21}{1000}\) cubic-dcm = 4.851. cubic dcm.

Hence the required number of sphere is 8 and the volume of each sphere is 4.851 cubic-dcm.

Example 6. The radius of base of a solid right circular iron-rod is 32 cm and the length of the rod is 35 cm. How many solid right circular cone of radius of base 8 cm and of height 28 Cm can be made by melting this rod?

Solution:

Given:

The radius of base of a solid right circular iron-rod is 32 cm and the length of the rod is 35 cm.

The radius of the base of the rod = 32 cm

and its length = 35 cm.

∴ volume of the rod = π x (32)2 x 35 Cc

Again, the radius of base of each cone = 8 cm

and height of each cone = 28 cm.

∴ volume = \(\frac{1}{3}\) x π x (8)2 x 28 cc

Let the number of cone be x.

∴ \(\frac{1}{3}\)π x (8)2 x 28 x x = π x (32)2 x 35

⇒ x =\(x=\frac{32 \times 32 \times 35 \times 3}{8 \times 8 \times 28} .\)

⇒ x = 60.

Hence the required number of solid cone is 60.

Class 10 Maths Solutions Wbbse

Example 7. The external radius of a hollow sphere of thickness 1 cm made of lead-plate is 6 cm. If a solid right circular rod of radius 2 cm is made by melting this hollow sphere, then what will be the length of the rod?

Solution:

Given:

The external radius of a hollow sphere of thickness 1 cm made of lead-plate is 6 cm. If a solid right circular rod of radius 2 cm is made by melting this hollow sphere

The external radius of the hollow sphere = 6 cm

The thickness of the sphere = 1 cm

∴ the inner radius of the hollow sphere = (6 – 1) cm = 5 cm

So, the volume of the materials of the sphere = \(\frac{4}{3}\)π(63 – 53) cc

= \(\frac{4}{3}\)π(216 -125) cc

= \(\frac{4}{3}\)π x 91 cc

The radius of the right circular rod = 2 cm.

Let the length of the rod = x cm

∴ the volume of the rod = π x (2)2 x x cc = 4πx cc

As per question, 4πx= \(\frac{4}{3}\) x π x 91

⇒ x =\(\frac{4 \times 91}{3 \times 4}\)

⇒ x = \(\frac{91}{3}=30 \frac{1}{3}\)

Hence the length of the solid right circular rod =30\(\frac{1}{3}\) cm

Example 8. What will be the ratio of the volumes of a solid cone, a solid hemisphere and a solid cylinder when their radii of bases and heights are equal?

Solution:

Let radius of the cone, hemisphere and cylinder be r unit and their equal heights be h unit.

Since the heights of the three solid objects are equal,

∴ height of the cone = height of the cylinder = height of the hemisphere = radius of the hemisphere = r unit.

∴ volume of the cone: volume of the hemisphere: volume of the cylinder

= \(\frac{1}{3} \pi r^2 h: \frac{2}{3} \pi r^3: \pi r^2 h\)

= \(\frac{1}{3} \pi r^2 \cdot r: \frac{2}{3} \pi r^3 \cdot: \pi r^2 \cdot r\)

= \(\frac{1}{3} \pi r^3: \frac{2}{3} \pi r^3: \pi r^3\)

= \(\frac{1}{3}: \frac{2}{3}: 1\)

= 1:2 : 3.

Hence the required ratio is 1 : 2 : 3.

Example 9. The upper part of a solid right circular cylindrical pillar is a hemisphere. What will be the volume of the pillar if the radius of base of it is 2 metres and its total length is 10 metres?

Solution:

Given:

The upper part of a solid right circular cylindrical pillar is a hemisphere.

Since the radius of base is 2 metres, so the radius of the hemisphere of the upper, part of the pillar is 2 metres.

∴ the height of the cylindrical part = (10 – 2) metres = 8 metres.

Now, the volume of the hemisphere =\(\frac{1}{2}\) :\(\frac{4}{3}\) x π x 23 cubic-metre

= \(\frac{16}{3}\)π cubic-metre

and the volume of the cylindrical part = π x (2)2 x 8 cubic-metre = 32π cubic-metre.

∴ the total volume = \(\left(\frac{16}{3} \pi+32 \pi\right)\)cubic-metre

= \(\pi\left(\frac{16}{3}+32\right)\)cubic-metre

= \(\pi \times \frac{112}{3}\)cubic-metre

= \(\frac{22}{7} \times \frac{112}{3}\)cubic-metre

= \(\frac{352}{3}\)cubic-metre

=117\(\frac{1}{3}\)cubic-metres

Hence the total volume of the pillar 117\(\frac{1}{3}\) cubic-metres.

Class 10 Maths Solutions Wbbse

Example 10. The lower part of a tent is of shape of a right circular cylinder, the height of which is 3.5 metres. The upper part of the tent is of shape of a right circular cone. If the total height of the tent be 9.5 metre and its diameter of base be 5 metre, then how much quantity of tarpaulin will be required to make 14 such tents?

Solution:

Given:

The lower part of a tent is of shape of a right circular cylinder, the height of which is 3.5 metres. The upper part of the tent is of shape of a right circular cone. If the total height of the tent be 9.5 metre and its diameter of base be 5 metre

The diameter of the base of the tent = 5 cm

∴ the radius of the base of the tent = \(\frac{5}{2}\) = 2.5 m

height of the cylindrical part = 3.5 m

∴ the curved surface area of this part = 2πrh = 2 x\(\frac{22}{7}\) x 2.5 x 3.5 sq-metre = 55 sq-metre

The radius of the conical upper part of the tent = 2.5 m and height = (9.5 – 3.5) m = 6 m.

The slant height of the conical part = \(\sqrt{r^2+h^2}\)

or, \(l=\sqrt{(2.5)^2+(6)^2} \mathrm{~m}\)

or, \(l=\sqrt{42.25} \mathrm{~m}\) = 65 m

∴ the curved surface area of this conical part = πrl

= \(\frac{22}{7}\) x 25 x 6.5 sq-m

∴ the tarpaulin required to make one tent = \(\left(55+\frac{22}{7} \times 2.5 \times 6.5\right)\)sq-metre

∴ the tarpaulin required to make 14 tent = 14 \(\left(55+\frac{22}{7} \times 2.5 \times 6.5\right)\) x 2.5 x 65 = 1485 sq-metre.

Hence the required quantity of tarpaulin = 1485 sq-metre.

Example 11. A conical flask of height 24 cm is full of water. If this water is poured into another cylindrical flask of radius half of the conical flask, then how much water-level of the cylindrical flask will be raised?

Solution:

Given:

A conical flask of height 24 cm is full of water. If this water is poured into another cylindrical flask of radius half of the conical flask

Let the radius of the base of the conical flask = r cm

∴ the radius of base of the cylindrical flask = \(\frac{r}{2}\) cm

∴ the volume of the conical flask = \(\frac{1}{3}\)πr2 cc

= \(\frac{1}{3}\)πr2 x 24 cc = 8πr2 cc

Let water-level into the cylindrical flask will rise h cm.

As per condition, π x \(\left(\frac{r}{2}\right)^2\) x h = 8πr2

\(\frac{1}{4}\)πr2h = 8πr2 ⇒ h = 32.

Hence the water-level into the cylindrical flask will rise 32 cm.

Example 12. One part of an Iron-pillar is cylimlrical and the rest part is conical. The radii of the cylindrical and conical part arc 8 cm and their heights are 240 cm and 36 cm respectively. If the weight of I ec iron he 7.8 gm, then what will be the total weight of the pillar?

Solution:

Given:

One part of an Iron-pillar is cylimlrical and the rest part is conical. The radii of the cylindrical and conical part arc 8 cm and their heights are 240 cm and 36 cm respectively. If the weight of I ec iron he 7.8 gm

The radius of the cylindrical part 8 cm and its height = 240 cm.

∴ the volume of the cylindrical part = \(\frac{22}{7}\) x 82 x 240 cubic cm

= \(\frac{22}{7}\)x 64 x 240 cc

Again, the radius of the conical part = 8 cm and its height = 36 cm.

∴ the volume of the conical part \(\frac{1}{3}\) x \(\frac{22}{7}\) x 82 x 36 cc

= \(\frac{22}{7}\) x 64 x 12 cc

So, the total volume of the pillar = \(\left(\frac{22}{7} \times 64 \times 240+\frac{22}{7} \times 64 \times 12\right)\) cc

= \(\frac{22}{7}\) x 64 x (240 + 12) cc

= \(\frac{22}{7}\) x 64 x 252 cc

Now, the weight of 1 cc pillar is 7.8 gm.

∴ the weight of \(\frac{22}{7}\) x 64 x 252 cc is 7.8 x \(\frac{22}{7}\) x 64 x 252 gm

= 395366.4 gm = 395.3664 kg

Hence the weight of the total pillar = 395.37 kg (approx.)

Example 13. Water flows at a speed of 10 m per minute through a right circular pipe of diameter 5 mm. In how much time a tank of conical shape having diameter 40 cm and depth 24 cm will be fulfilled completely by this pipe?

Solution:

Given:

Water flows at a speed of 10 m per minute through a right circular pipe of diameter 5 mm.

The diameter of the conical tank = 40 cm

∴ its radius = \(\frac{40}{2}\) cm = 20 cm

The depth of the tank = 24 cm

∴ the volume of the tank = \(\frac{1}{3}\) x π x (20)2 x 24 cc

= \(\frac{1}{3}\)π x x 20 x 20 x 24 cc = 3200π cc

Let the tank will be fulfilled completely in x minute.

∴ the length of the water-column = 10x metres = 1000x cm

Clearly, the length of the cylindrical water-column is 1000x cm

and its radius = \(\frac{5}{2}\) mm = \(=\frac{5}{2 \times 10}\) cm = \(\frac{1}{4}\) cm

∴ the volume of cylindrical water-column = \(\pi \times\left(\frac{1}{4}\right)^2 \times 1000 x \mathrm{cc}\)

= \(\pi \times \frac{1}{16} \times 1000 x \mathrm{cc}\)

As per question, \(\frac{\pi \times 1000 x}{16}\) = 3200π

⇒ x = \(\frac{3200 \times 16}{1000}=\frac{256}{5}=51 \frac{1}{5}\)

Hence the tank will be completely fulfilled in 51\(\frac{1}{5}\) minutes.

Class 10 Maths Solutions Wbbse

Example 14. A pot of conical shaped, the radius and height of which is 6 cm and 8 cm respectively, is fulfilled with water. Now, a sphere is completely immersed into the water of this pot in such a wayn that it just touches the two sides of the pot. So, how many part of the water of the pot will over flow?

Solution:

Given:

A pot of conical shaped, the radius and height of which is 6 cm and 8 cm respectively, is fulfilled with water. Now, a sphere is completely immersed into the water of this pot in such a wayn that it just touches the two sides of the pot.

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Question Example 14

 

AD = 6 cm and DC = 8 cm

∴ AC = \(\sqrt{C D^2+A D^2}\)

= \(\sqrt{8^2+6^2} \mathrm{~cm}\)

= \(\sqrt{64+36} \mathrm{~cm}\)

= 10 cm

Now, in the right-angled triangles ΔACD and ΔEOC, we get, ∠ACD = ∠ECO

Let the radius of the sphere be x cm.

∴ AD = AE = 6 cm

∴ EC = (AC – AE) = (10 – 6) cm = 4 cm

Again, ΔADC ~ ΔEOC,

∴ \(\frac{\mathrm{DC}}{\mathrm{AD}}=\frac{\mathrm{EC}}{\mathrm{EO}}\)

⇒ \(\frac{8}{6}=\frac{4}{x}\)

⇒ \(x=\frac{4 \times 6}{8}=3\)

∴ the radius of the sphere is 3 cm

So, the volume of the sphere = \(\frac{4}{3}\)π x 33 cc = 3671 cc

Again, the volume of the conical pot = \(\frac{1}{3}\)π x 62 x 8 cc= 96π cc

So, the required part = \(\frac{36 \pi}{96 \pi}=\frac{3}{8}\)

Hence \(\frac{3}{8}\) part of the water of the pot will overflow.

Example 15. The lower part of a tent is a right circular cylinder of radius 14 metres and its upper part is a right circular cone. If the height of the cylindrical part is 3 metre and the total height of the tent is 13.5 metre, then find the expenditure of colouring the inner part of the tent at a rate of 2 per square-metre.

Solution:

Given:

The lower part of a tent is a right circular cylinder of radius 14 metres and its upper part is a right circular cone. If the height of the cylindrical part is 3 metre and the total height of the tent is 13.5 metre

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Question Example 15

 

The radius of the cylindrical part = 1 4 m and the height of this part = 3 m.

∴ the curved surface area of the cylindrical part = 2 x \(\frac{22}{7}\)x 14 x 3 square-metre = 264 square-metre

Again, the height of the conical part of the tent = (13.5 – 3) m = 10.5 m

and its radius of base = 14 metre

∴ the slant height of the conical part

= \(\sqrt{(14)^2+(10.5)^2}\) metre

= \(\sqrt{196+110.25}\) metre

= \(\sqrt{306.25}\) metre

= 17.5 metre

∴ the curved surface area of the conical part of the tent = \(\frac{22}{7}\) x 14 x 17.5 sq-metre = 770 square-metre

∴ the area of the part of the tent which is to be coloured = (770 + 264) square-metre = 1034 square-metre

So the expenditure of colouring = ₹2 x 1034 = ₹2068

Hence the required expenditure = ₹2068.

Class 10 Maths Solutions Wbbse

Example 16. The radii of the two ends of a pail of height 24 cm are 15 cm and 5 cm. Find the volume of the pail.

Solution:

Given:

The radii of the two ends of a pail of height 24 cm are 15 cm and 5 cm.

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Question Example 16

 

The volume of the pail = (Volume of the conical OAB) – (Volume of. the cone OCD).

Now, ΔOPB ~ ΔOQD,

∴ \(\frac{\mathrm{PO}}{\mathrm{OQ}}=\frac{\mathrm{PB}}{\mathrm{QD}}\)

⇒ \(\frac{h_1}{h_2}=\frac{15}{5}=3\)

⇒ h1 = 3h2

Again, PQ = OP – OQ

⇒ h1 -h2 = 24

⇒ 3h1 – h2 = 24

⇒ 2h2 = 24

⇒ h2 = \(\frac{24}{2}\)

⇒ h2 = 12

∴ h1 = 3h2 = 3 x 12 = 36

So, the volume of the cone OAB = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 152 x 36 cc

= \(\frac{22}{7}\) x 5 x 15 x 36 cc

and the volume of the cone OCD = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 152 x 12 cc

= \(\frac{22}{7}\) x 5 x 5 x 4 cc

∴ the volume of the pail = \(\left(\frac{22}{7} \times 5 \times 15 \times 36-\frac{22}{7} \times 5 \times 5 \times 4\right) \mathrm{cc}\)

= \(\frac{22}{7} \times 5(15 \times 36-5 \times 4) \mathrm{cc}\)

= \(\frac{22}{7} \times 5(540-20) \mathrm{cc}=\frac{22}{7} \times 5 \times 520 \mathrm{cc}\)

= \(\frac{57200}{7} \mathrm{cc}=817.1 \frac{3}{7} \mathrm{cc}\)

Hence the volume of the pail= 8171 \(\frac{3}{7} \mathrm{cc}\) cc.

Class 10 Maths Solutions Wbbse

Example 17. A hemisphere of internal diameter 36 cm is filled with water. If this water is poured into right circular bottles of radius 3 cm and of height 6 cm, then how many bottles will be needed?

Solution:

Given:

A hemisphere of internal diameter 36 cm is filled with water. If this water is poured into right circular bottles of radius 3 cm and of height 6 cm

The radius of the hemisphere = \(\frac{36}{2}\)cm = 18 cm

∴ the volume of the hemisphere = \(\frac{2}{3}\) x π x 183 cc

The radius of each right circular bottle = 3 cm and its height = 6 cm.

∴ The volume of each bottle = π x 32 x 6 cc

Let the number of bottles to be needed be x.

∴ x x π x 32 x 6 = \(\frac{2}{3}\) x π x 183.

⇒ x = \(\frac{2 \times 18 \times 18 \times 18}{3 \times 3 \times 3 \times 6}\)

⇒ x = 72

Hence the required number of bottles = 72.

Example 18. The ratio of the volumes of a cylinder and a sphere is 1 : 3 and the ratio of the radii of the base of the cylinder and of the sphere is 1 : 3. If the sum of the height and radius of base of the cylinder is 78 cm, then what will be the height of the cylinder?

Solution:

Given:

The ratio of the volumes of a cylinder and a sphere is 1 : 3 and the ratio of the radii of the base of the cylinder and of the sphere is 1 : 3. If the sum of the height and radius of base of the cylinder is 78 cm

Let the radii of the cylinder and the sphere be x cm and 3x cm respectively.

Also let the volumes of the cylinder and the sphere be V and 3V respectively.

As per question, x + h = 78 ………..(1)

Again, V = πx2h cu-unit and 3V = \(\frac{4}{3}\)π x (3x)3 cu-unit

⇒ 3πx2h = 36πr3 ⇒ h = \(\frac{36 x^3}{3 x^2}\) = 12x

∴ from (1) we get, 12x + x = 78

or, 13x = 78

or, x = \(\frac{78}{13}\) = 6

and h = 12x

⇒ h = 12 x 6 = 72.

Hence the height of the cylinder = 72 cm.

Class 10 Maths Solutions Wbbse

Example 19. The total surface area of a solid hemisphere is 1848 sq-cm. A solid right circular cone is made by melting this hemisphere. If the radius of base of the cone is equal to the radius of the hemisphere, then what will be the height of the cone?

Solution:

Given:

The total surface area of a solid hemisphere is 1848 sq-cm. A solid right circular cone is made by melting this hemisphere. If the radius of base of the cone is equal to the radius of the hemisphere

Let the radius of the hemisphere be r cm.

So, the total surface area of the hemisphere = 3πr2sq-cm

As per the question, 3πr2 = 1848

or, 3 x \(\frac{22}{7}\) x r2 = 1848

or, r2 = \(\frac{1848 \times 7}{3 \times 22}\)

or, r2 = 196

Also, the volume of the hemisphere = \(\frac{2}{3}\)πr3 cc

and the volume of the cone = \(\frac{1}{3}\)πr2hcc

As per question, \(\frac{2}{3}\)πr3 = \(\frac{1}{3}\)πr2h

⇒ h = \(\frac{2 \pi r^3}{\pi r^2}\)

⇒ h = 2r

⇒ h = 2 x 14

⇒ h = 28

Hence the height of the cone = 28 cm.

Example 20. Half of a tank of length 5 metre, breadth 4 metre and height 2.2 metre, is filled with water. How many iron-bullets of radius 5 cm each should be completely immersed into the water of the tank so that the tank will full to the brim?

Solution:

Given:

Half of a tank of length 5 metre, breadth 4 metre and height 2.2 metre, is filled with water.

The radius of each iron bullet = 5 cm

∴ the volume of each iron-bullet = \(\frac{4}{3}\)π x 53 cc

Let the required number of iron-bullets be x.

the total volume of x bullets = x x \(\frac{4}{3}\)π x 53 cc .

Since half of the tank is filled with water, the Water-level of the tank will be increased \(\frac{2.2}{2}\) m = 1.1 m to be filled to the brim.

So, the volume of the raised-water = 5 x 4 x 1.1 cubic-metre

= 500 x 400 x 110 cc

As per question, \(\frac{4}{3}\) x \(\frac{22}{7}\) x 53 x x = 500 x 400 x 110

⇒ x = \(\frac{500 \times 400 \times 110 \times 3 \times 7}{4 \times 22 \times 125}\)

⇒ x = 42000

Hence the required number of iron-bullets = 42000.

Example 21. Half of a tank of length 21 dcm, breadth 11 dcm and depth 6 dcm is full of water. If 100 spheres made of iron and of diameter 21 cm each are fully immersed into this water of tank, then how many dcm of the water-level of the tank will be raised?

Solution:

Given:

Half of a tank of length 21 dcm, breadth 11 dcm and depth 6 dcm is full of water. If 100 spheres made of iron and of diameter 21 cm each are fully immersed into this water of tank

The length of the tank = 21 dcm

breadth of the tank =11 dcm; and

depth of the tank = 6 dcm.

Let the water-level of the tank rises x dcm after the complete immersion of 100 iron-spheres.

∴ the volume of the raised-water = (21 x 11 x x) cubic-dcm.

Again, diameter of each sphere = 21 cm

∴ radius of each sphere = \(\frac{21}{2}\) cm = \(\frac{21}{2 \times 10}\) x 10 = 1.05 dcm.

∴ the volume of 100 iron-spheres = 100 x \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.05)3 cubic-dcm

As per question, 21 x 11 x x = 100 x \(\frac{4}{3}\) x \(\frac{22}{7}\) x (1.05)3

x =\(\frac{100 \times 4 \times 22 \times 1.05 \times 1.05 \times 1.05}{3 \times 7 \times 21 \times 11}\)

= \(\frac{100 \times 4 \times 22 \times 105 \times 105 \times 105}{3 \times 7 \times 21 \times 11 \times 100 \times 100 \times 100}\)

= 2.1

Hence the water-level of the tank will rise 2. 1 dcm.

Class 10 Maths Solutions Wbbse

Example 22. The cross-section of a rectangular parallelopiped wooden log of length 2 metre is squareshaped and each of its side is of length 14 dcm. If the rectangular log is reduced to a right circular log by loosing smallest amount of wood, then how much cubic-metre of wood the log will posses and how cubic-metre of wood will be lost?

Solution:

Given:

The cross-section of a rectangular parallelopiped wooden log of length 2 metre is squareshaped and each of its side is of length 14 dcm. If the rectangular log is reduced to a right circular log by loosing smallest amount of wood

WBBSE Solutions For Class 10 Maths Mensuration Chapter 5 Problems Related To Different Solids And Objects Long Answer Question Example 22

 

The cross-section of the rectangular parallelopiped wooden log is square-shaped and each of its side is of length 14 dcm.

The rectangular log is reduced to a circular log.

∴ the length of the diameter of the circular area = the length of the side of the square.

∴ the radius ofthe right circular cylindrical wooden

log = \(\frac{14}{2}\) dcm = 7 dcm.

The length of the log = 2 metre = 2 x 10 dcm = 20 dcm

∴ the volume of the rectangular wooden log= 14 x 14 x 20 cubic-dcm

= 3920 cubic-dcm = 3.92 cubic-metre.

Again, the volume of the circular log = \(\frac{22}{7}\) x 72 x 20 cubic-dcm

= 3080 cubic-dcm = 3.08 cubic-metre

∴ the lost wood = (3.92 – 3.08) cubic-metre.

= 0.84 cubic-metre

Hence the right circular cylindrical log will possess 3.08 cubic-metre of wood and the lost wood is 0.84 cubic-metre.

Example 23. What will be the ratio of the volumes of a sphere and a cube if the surface areas of the sphere and the cube are equal?

Solution:

Let the radius of the sphere = r unit

and let the length of each side of the cube = x unit.

As per question, 4πr2 = 6x2

⇒ \(\frac{r^2}{x^2}=\frac{3}{2 \pi}\)

⇒ \(\frac{r}{x}=\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{\pi}}\)

∴ (volume of the sphere) : (volume of the cube)

= \(\frac{4}{3} \pi r^3: x^3=\frac{\frac{4}{3} \pi r^3}{x^3}=\frac{4}{3} \pi\left(\frac{r}{x}\right)^3\)

= \(\frac{4}{3} \times \pi \times\left(\frac{\sqrt{3}}{\sqrt{2} \cdot \sqrt{\pi}}\right)^3=\frac{4}{3} \times \pi \times \frac{3 \sqrt{3}}{2 \sqrt{2} \pi \sqrt{\pi}}\)

= \(\frac{\sqrt{2} \times \sqrt{3}}{\sqrt{\pi}}=\frac{\sqrt{6}}{\sqrt{\pi}}=\sqrt{6}: \sqrt{\pi}\)

Hence the required ratio =√6 : √π.

Example 24. A right circular cone*is made by cutting a cubical wooden block of side 9 cm each, then what will be the greatest Volume of the cone?

Solution:

Given:

A right circular cone*is made by cutting a cubical wooden block of side 9 cm each

The length of each side of the cube = 9 cm

i.e., if the radius of the required cone be r cm, then 2r = 9 ⇒ r = \(\frac{9}{2}\)

∴ the height of the cone = 9 cm

∴ Volume of the cone = \(\frac{1}{3}\)πr2h cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\left(\frac{9}{2}\right)^2\) x 9 cc

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x \(\frac{9}{2}\) x \(\frac{9}{2}\) x 9 cc

= \(\frac{2673}{14}\) cc

= 190.93 cc (approx.)

Hence the greatest volume of the cone = 190.93 cc (approx.)

 

 

 

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone

Mensuration Chapter 4 Right Circular Cone

What is a right circular cone?

If a right-angled triangle is rotated by taking one of its sides as the axis except the hyponetuse with ah angle of 360°, then the solid object thus produced is known as right circular cone.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Right Circular Cone

 

For example, a comet of Muri, a cone of ice cream, etc.

The right-angular point of the right-angled triangle is O and ∠AOB = 90°.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Right Circular Cone Is Rotated

 

Now, keeping the right-angular point O fixed and taking AO as the axis if the triangle is rotated with an angle of 360°, then the right circular angle ABC is produced.

The line segment joining the vertex and the centre of the circular base will be perpendicular to the base of the cone.

The line segment joining the vertex and the centre of the circular base is not perpendicular to the base of the cone.

WBBSE Solutions for Class 10 Maths

So it is not a right circular cone.

Remember that a closed right circular cone has two surfaces-one is the curved surface and the other is the plane surface.

Some essential definition regarding a right circular cone:

Axis: The side, of a right-angled triangle adjacent to the right angle around which the triangle is rotated with an angle of 360°, a right circular cone is produced is biown as the axis of the cone.

The axis of the right circular cone is AO.

Base: The circular plane, which is produced by the other side of the right-angled triangle adjacent to the right angle, when the triangle is rotated with an angle of 360°, is known as the base of the right-circular cone.

Maths Solutions Class 10 Wbbse

The base of a right circular cone is a circle. O is the centre of the circular base.

Vertex: The point, at the top of the right circular cone which remains fixed while producing the cone by rotating the right-angled triangle with jan angle of 360° except the right-angular point is known as the vertex of the right circular cone.

A is the vertex of the right circular cone.

The radius of the base: The base of the right circular cone is-a circle and the radius of this circle is known as the radius of the base.

OB is the radius of the base of the right-circular cone.

Height: The perpendicular distance between the centre of the base of the right circular cone and the vertex of the cone is known as the height of the right circular cone.

AO is the height of the right circular cone AOB.

Slant height: The length of the hypotenuse of the right-angled triangle which by rotation produces the cone is known as the slant height of the right circular cone.

AB is the slant height of the right circular cone.

Semi-vertical angle: The angle between the axis of a right circular cone and its slant height is known as the semi-vertical angle.

∠OAB is the semi-vertical angle.

Relation among the slant height, height and the radius of the base of a right circular cone:

Let ABC be the right circular cone produced by rotating the right-angled triangle AOB by keeping the side AO fixed and with an angle of 360°.

Here, the height of the cone = AO and the radius of it = OB. The slant height of the cone = AB.

Maths Solutions Class 10 Wbbse

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Radius Of the Base Of A Right Circular Cone

 

∴ by Pythagoras theorem, AB2 = OA2 + OB2 i.e.,

(slant height)2 = (height)2 + (radius of the base)2

Hence if the slant height, height and the radius of the base be l, h and r Unit respectively, then l2 = h2 + r2

or, l = \(\sqrt{h^2+r^2}\) and r = \(\sqrt{h^2+r^2}\) and h = \(\sqrt{l^2-r^2}\)

Formulas related to right circular cone:

Let the radius of the base of a right circular cone is r unit, height h unit and slant height l unit. Then

1. The lateral surface area of the right circular cone

= \(\frac{1}{2}\) x circumference of the circular base x slant height = \(\frac{1}{2}\) x 2πr x l sq-unit = πrl sq-unit

2. The area of the base of the right circular cone = area of the circular base = πr2 sq-unit.

3. The total surface area of the right circular cone = lateral surface area + area of the base = (πrl + πr2) sq-unit = πr (l + r) sq-unit.

4. The volume of the right circular cone = \(\frac{1}{3}\)πr2h cubic-unit.

 

Mensuration Chapter 4 Right Circular Cone Multiple Choice Questions

Example 1. The ratio of the volumes of two right circular cones is 1: 4 and the ratio of their lengths of radii of the bases is 4: 5. Then the ratio of their heights is

  1. 1: 5
  2. 5: 4
  3. 25: 16
  4. 25: 64

Solution: Let the lengths of the radii of the bases of the right circular cone are 4x unit and 5x unit.

Also, let the heights of the two right circular cones are h1 unit and h2 unit.

So, the volume of the cones are \(\frac{1}{3}\)π(4x)2 h1 cubic-units and \(\frac{1}{3}\)π(5x)2 h2 cubic-unit.

As per question, \(\frac{1}{3}\)π(4x)2 h1 : \(\frac{1}{3}\)π(5x)2 h2 cubic-unit.

⇒ 16 x2 h1 : 25 x2 h2 = 1: 4

⇒ \(\frac{16 x^2 h_1}{25 x^2 h_2}=\frac{1}{4}\)

⇒ \(\frac{h_1}{h_2}\) = \(\frac{1}{4}\times \frac{25}{16}\)

⇒ \(\frac{h_1}{h_2}=\frac{25}{64}\)

⇒ h1: h2 = 25: 64

Hence the ratio of the heights of the two right circular cones is 25: 64.

∴ 4. 25: 64 is correct.

The ratio of their heights is  25: 64

Maths Solutions Class 10 Wbbse

Example 2. If keeping the length of radius of a right circular cone fixed, the height of it is increased by 2 times, then the volume of cone will be increased by

  1. 100 %
  2. 200 %
  3. 300 %
  4. 400 %

Solution: Let the radius of the right circular cone be r unit and height is h unit.

∴ if the volume of the cone be V cubic-unit, then V = \(\frac{1}{3}\)πr2h.

Now, if the height of the cone be doubled keeping, the radius of it fixed, then the new volume of the cone

V1 = \(\frac{1}{3}\)πr2 (2h) = 2. \(\frac{1}{3}\)πr2h = 2v…….(1)

So, the volume increases = (V1 – V) cubic-unit = (2V – V) cubic-unit = V cubic-unit.

Hence the percent of the increment in volume

= \(\frac{V}{V}\) x 100%\(\left[\frac{\text { increment in volume }}{\text { primary volume }} \times 100 \%\right]\) = 100%

∴ 1. 100% is correct

The volume of cone will be increased by 100%

Wbbse Class 10 Maths Solutions

Example 3. If each of the radius and height of a right circular cone be doubled, then the volume of the cone will be

  1. 3 times
  2. 4 times
  3. 6 times
  4. 8 times

Solution: If the radius, height and volume of the cone be r unit, h unit and V cubic-unit respectively, then V = \(\frac{1}{3}\)πr2h .

Now, if the volume of the cone be V1 cubic-unit when its radius and height be doubled, then

V1 = \(\frac{1}{3}\)π(2r)2.2h = \(\frac{1}{3}\)π.4r2 .2h = 8.\(\frac{1}{3}\)πr2 h = 8V

∴ If the radius arid height of a right circular cone be doubled its volume will be 8 times of the previous volume.

∴ 4. 8 times is correct.

The volume of the cone will be 8

Example 4. If the length of radius of a right circular cone be \(\frac{r}{2}\) unit and its slant height be 2l unit, the total surface area of it will be

  1. 2πr (l + r) sq-unit
  2. πr (l + r) sq-unit
  3. \(\pi r\left(1+\frac{r}{4}\right)\) sq-unit
  4. 2πrl sq-unit

Solution: The radius of the right circular cone is \(\frac{r}{2}\), slant height = 2l unit

∴ the total surface area of the cone = \(\pi \times \frac{r}{2}\left(\frac{r}{2}+2 l\right)\)sq- unit

= \(\pi \times \frac{r}{2} \times 2\left(\frac{r}{4}+l\right)\)sq- unit

= \(\pi r\left(l+\frac{r}{4}\right)\) sq-unit

∴ 2. πr (l + r) sq-unit is correct.

The total surface area of it will be πr (l + r) sq-unit

Example 5. If a triangle of sides 6 cm, 8 cm and 10 cm be rotated keeping the side of length 8 cm fixed, then the volume of the cone thus produced will be

  1. 96 π cc
  2. 120 π cc
  3. 128 π cc
  4. 200 π cc

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Triangle Of Sides

 

Solution: Since 62 + 82 = 102,

∴ the given triangle is a right-angled triangle, the hypotenuse of which is 10 cm.

Now if the triangle is rotated keeping the side of length 8 cm fixed, then height of the cone will be 8 cm and the radius of its base will be 6 cm

[∵ the other side of length 10 cm is the hypotenuse]

∴ the volume of the cone = \(\frac{1}{3}\)π x 62 x 8 cc = 96π cc

∴ 1. 96 π cc is correct.

The volume of the cone thus produced will be 96 π cc

Example 6. The area of the base of a right-circular cone is 25 sq-cm and the height is 10 cm. Then the volume of the cone will be

  1. \(\frac{150}{3}\) cc
  2. \(\frac{250}{3}\) cc
  3. \(\frac{350}{3}\) cc
  4. \(\frac{400}{3}\) cc

Solution: The volume of the cone = \(\frac{1}{3}\) x area of the base x height = \(\frac{1}{3}\) x 25 x 10 cc = \(\frac{250}{3}\) cc

∴ 2. \(\frac{250}{3}\) cc is correct.

The volume of the cone will be \(\frac{250}{3}\) cc

Example 7. If the numerical values of the area of the base and the volume of a right circular cone of radius 4 cm be equal, then the slant height of the cone is

  1. 3 cm
  2. 4 cm
  3. 5 cm
  4. 6 cm

Solution: Let the height of the cone be h cm.

As per question, \(\frac{1}{3}\)π .42 .h = π . 42 ⇒ h = 3

∴ the slant height of the cone = \(\sqrt{4^2+3^2}\)

∴ 3. 5 cm is correct.

The slant height of the cone is 3. 5 cm

Wbbse Class 10 Maths Solutions

Example 8. If the height of a right circular cone be increased by 10% keeping its radius fixed, then the volume of the cone will be

  1. 5%
  2. 10%
  3. 15%
  4. 20%

Solution: Let the radius of the base of a right circular cone be r unit and its height be h unit.

∴ if the volume of the cone be V cubic-unit, then V = \(\frac{1}{3}\)πr2h…….(1)

Now, if the height be increased by 10% then the height will be

\(\left(h+h \times \frac{10}{100}\right)\) unit = \(\left(h+\frac{h}{10}\right)\) unit = \(\frac{11 h}{10}\) unit

 

Then the volume of the cone = \(\frac{1}{3}\)πr2 x \(\frac{11 h}{10}\) cubic-unit = \(\frac{11 h}{10}\) x \(\frac{1}{3}\)πr2h cubic -unit

= \(\frac{11 V}{10}\) cubic -unit [by(1)]

∴ the icrement of volume = \(\left(\frac{11 \mathrm{~V}}{10}-\mathrm{V}\right)\) cubic- unit = \(\frac{V}{10}\) cubic- unit

So, the percent of increment of volume = \(\frac{\frac{\mathrm{V}}{10}}{\mathrm{~V}}\) x 100% = 10%

∴ 2. 10% is correct.

The volume of the cone will be 10%

Example 9. The volume of a right circular conical tent is 1232 cc. If the height of the tent be 24 metres, the area of the base of the cone will be

  1. 140 sq-m
  2. 145 sq-m
  3. 154 sq-m
  4. 160 sq-m

Solution: Let the radius ofthe base of the tent be r metre.

∴ the volume of the tent = \(\frac{1}{3}\)πr2 x 24 cc

∴ \(\frac{1}{3} \times \frac{22}{7} \times r^2\) x 24 = 1232

or, \(r^2=\frac{1232 \times 3 \times 7}{22 \times 24}\)

∴ the area of the base of tent = nr2 sq-cm = \(\frac{22}{7} \times \frac{1232 \times 7 \times 3}{22 \times 24}\) sq-cm = 154 sq-cm.

∴ 3. 154 sq-m is correct.

The area of the base of the cone will be 154 sq-m

Wbbse Class 10 Maths Solutions

Example 10. The difference of the square of the height and the square of the base of a right circular cone of slant height 10 cm is 28 cm, then the volume of the cone will be

  1. 42 π cc
  2. 54 π cc
  3. 72 π cc
  4. 96 π cc

Solution: Let the height of the right circular cone is h cm and the radius of the base be r cm.

The slant height of the cone is 10 cm.

∴ h2 + r2 = (10)2 or, h2 + r2 = 100…..(1)

Again, h2 – r2 = 28…….(2)

Adding (1) and (2) we get, 2h2 = 128 or, h2 = 64 or, h = 8.

r2 = 100 – h2 = 100 – 64 = 36.

∴ the volume of the cone = \(\frac{1}{3}\)π x n x 36 x 8 cc = 96πcc.

∴ 4. 96 cc is correct.

The volume of the cone will be 96 cc.

 

Mensuration Chapter 4 Right Circular Cone True Or False

Example 1. If the radius of the base of a right circular cone be halved and its height be doubled, then the volume of the cone remains the same.

Solution:

Given:

If the radius of the base of a right circular cone be halved and its height be doubled

If the volume of a right circular cone of radius r unit and height h unit be V cubicunit, then V = \(\frac{1}{3}\)πr2h

Now, if the radius of its base be halved and its height be doubled, then the volume of the cone

= \(\frac{1}{3} \pi\left(\frac{r}{2}\right)^2\).2h cubic -unit

= \(\frac{1}{3} \pi \cdot \frac{r^2}{4}\).2h cubic- unit

= \(\frac{1}{2}\).\(\frac{1}{3}\)πr2h cubic- unit = \(\frac{1}{2}\) V

i.e., the volume is halved.

Hence the given statement is false.

Example 2. The slant height of a right circular cone is always greater than the height of the cone.

Solution:

The height, radius and slant height of right circular cone are the three sides of a rightangled triangle.

Since slant height is the hypotenuse of this right-angled triangle, so it is always greater than its height.

Hence the given statement is true.

Wbbse Class 10 Maths Solutions

Example 3. Among the total surface area, lateral surface area and the area of the base of a right circular cone, the total surface area is the greatest in magnitude.

Solution:

Given:

Among the total surface area, lateral surface area and the area of the base of a right circular cone

Since for a right circular cone,

the total surface area = (area of lateral surface) + (area of base),

so, the total surface area is the greatest.

Hence the given statement is true.

Example 4. If the area of the base of a right circular cone be 3 times of its volume, then the height of it will be 1 unit.

Solution:

Given:

If the area of the base of a right circular cone be 3 times of its volume

Let the radius of the right circular cone be r unit and its height is h unit.

As per question, πr2 =3 x \(\frac{1}{3}\)πr2h ⇒ h = 1

∴ the height of the cone = 1 unit.

Hence the given statement is true.

Example 5. If the area of the base of a right circular cone be x sq-unit and the volume be y cubic-unit, then the height of the cone will be \(\frac{3x}{y}\) unit.

Solution:

Given:

If the area of the base of a right circular cone be x sq-unit and the volume be y cubic-unit

We know, volume = \(\frac{1}{3}\) x area of base x height

=» height = \(\frac{\text { volume } \times 3}{\text { area of base }}\)

= \(\frac{y \times 3}{x}\)unit = \(\frac{3 y}{x}\)unit.

Hence if the area of the base be x sq-unit and its volume be y cubic-unit, the height will be \(\frac{3 y}{x}\) unit.

Hence the given statement is true.

 

Mensuration Chapter 4 Right Circular Cone Fill In The Blanks

Example 1. AC is the hypotenuse of a right-angled triangle ABC. The radius of the right circular cone which is produced by one complete rotation of the triangle with the side AB as the axis is _______

Solution:

Given:

AC is the hypotenuse of a right-angled triangle ABC. The radius of the right circular cone which is produced by one complete rotation of the triangle with the side AB as the axis i

BC

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Hypotenuse Of Right Angled Triangle

 

Since the hypotenuse of the right-angled triangle is AC and the right circular cone is produced by one complete rotation of the triangle keeping the side AB fixed, so AB is the height of the cone.

Clearly, the radius of the base is BC.

Example 2. If the volume of a right circular cone be V cubic-unit and the area ofthe base be A sq-unit, then the height of the cone is ______

Solution:

Given:

If the volume of a right circular cone be V cubic-unit and the area ofthe base be A sq-unit

\(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

Since volume = \(\frac{1}{3}\) x area of the base x height

∴ V = \(\frac{1}{3}\) x A x h

⇒ h = \(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

The height of the cone is = \(\frac{3 \mathrm{~V}}{\mathrm{~A}}\)

Example 3. The number of surfaces of a closed right pircular cone is ______

Solution: 2

Example 4. The number of vertices of a right circular cone is ______

Solution: 1

Example 5. The lateral surface area of a right circular cone = (Total surface area) – ________

Solution: area of base

 

Mensuration Chapter 4 Right Circular Cone Short Answer Type Questions

Example 1. The height of a right circular cone is 12 cm and its volume is 100 n cc, then find the radius of the cone.

Solution:

Given:

The height of a right circular cone is 12 cm and its volume is 100 n cc

Let the radius of the right circular cone be r cm and its height is 12 cm.

∴ the volume of the cone = \(\frac{1}{3}\)πr2 x 12cc = 4πr2cc

As per question, 4πr2 = 100π or, r2 = 25, or, r = 5

Hence the radius of the cone = 5 cm.

Example 2. The curved surface area of a right circular cone is V5 times of its base area. Find the ratio of the height and the length of radius of the cone.

Solution:

Given:

The curved surface area of a right circular cone is V5 times of its base area.

Let the radius of the cone is r unit, height is h unit and slant height is l unit.

∴ the curved surface area of the cone = πrl sq-unit = πr\(\sqrt{h^2+r^2}\) sq- unit.

Also, the area of the base = πr2 sq-unit.

As per question, πr\(\sqrt{h^2+r^2}\) = √5πr2 or, \(\sqrt{h^2+r^2}\) = √5r or, h2 + r2 = 5r2 or, h2 = 4r2

or, h2 = (2r)2 or, h = 2r or, \(\frac{h}{r}=\frac{2}{1}\) or, h : r = 2 : 1.

Hence the ratio of the height and radius of the right circular cone is 2: 1.

Example 3. If the volume of a right circular cone is V cubic-unit, base area is A sq-unit and height is H unit, then find the value of \(\frac{AH}{V}\)

Solution:

Given:

If the volume of a right circular cone is V cubic-unit, base area is A sq-unit and height is H unit

Let the radius of the right circular cone is r unit. the volume of the cone,

V = \(\frac{1}{3}\)πr2H cubic- unit,

Again, the area of the base of the cone, A = πr2 sq-unit

∴ \(\frac{\mathrm{AH}}{\mathrm{V}}=\frac{\pi r^2 \mathrm{H}}{\frac{1}{3} \pi{ }^2 \mathrm{H}}\)

⇒ \(\frac{\mathrm{AH}}{\mathrm{V}}=3\)

Hence the value of \(\frac{AH}{V}\) = 3

Example 4. The numerical values of the volume and the lateral surface area of a right circular cone are equal. If the height and the radius of the cone are h unit and r unit respectively, then find the value of \(\frac{1}{h^2}+\frac{1}{r^2}\).

Solution:

Given:

The numerical values of the volume and the lateral surface area of a right circular cone are equal. If the height and the radius of the cone are h unit and r unit respectively

Let the slant height of the right circular cone is l unit.

Also, the height and radius of the cone are h and r unit respectively.

As per question, \(\frac{1}{3}\)πr2h = πrl

or, rh = 3l

⇒ r2h2  = 9l2  

⇒ r2 h2 = 9 (h2 + r2)

⇒ \(\frac{h^2+r^2}{r^2 h^2}=\frac{1}{9}\)

⇒ \(\frac{h^2}{r^2 h^2}+\frac{r^2}{r^2 h^2}=\frac{1}{9}\)

⇒ \(\frac{1}{r^2}+\frac{1}{h^2}=\frac{1}{9}\)

⇒ \(\frac{1}{h^2}+\frac{1}{r^2}=\frac{1}{9}\)

Hence the value of \(\frac{1}{h^2}+\frac{1}{r^2}=\frac{1}{9}\)

Example 5. The ratio of the lengths of the radii of the bases of a right circular cylinder and of a right circular cone is 3: 4 and the ratio of their heights is 2 : 3; Find the ratio of the volumes of the cylinder and the cone.

Solution:

Given:

The ratio of the lengths of the radii of the bases of a right circular cylinder and of a right circular cone is 3: 4 and the ratio of their heights is 2 : 3

Let the radii of the bases of the right circular cylinder and the right circular cone are 3r units and 4r units.

Also, let the heights of the cylinder and, the cone are 2h units and 3h units.

∴ the ratio of the volumes of the right circular cylinder and the right circular cone will be

\(\pi(3 r)^2 \cdot 2 h: \frac{1}{3} \pi(4 r)^2 \cdot 3 h=\pi \cdot 9 r^2 \cdot 2 h: \frac{1}{3} \pi \cdot 16 r^2 \cdot 3 h\)

= \(\frac{18 \pi r^2 h}{16 \pi r^2 h}\) = 18: 16 = 9: 8

Hence the required ratio is 9 : 8.

Example 6. The ratio of the radius of the base of a right circular cone and its height is 3: 7; The volume of the cone is 528 cc. Find the diameter of the cone.

Solution:

Given:

The ratio of the radius of the base of a right circular cone and its height is 3: 7; The volume of the cone is 528 cc.

Let the radius of the right circular cone and its height be 3x cm and 7x cm.

.-. the volume-of the cone = \(\frac{1}{3}\)π(3x)2.7x cubic- unit

As per question, \(\frac{1}{3}\)π(3x)2 . 7x = 528 or, \(\frac{1}{3}\)π x 9x2 . 7x = 528

or, \(\frac{1}{3} \times \frac{22}{7} \times 63 x^3\) = 528

or, x3 = \(=\frac{528 \times 3 \times 7}{22 \times 63}\) or, x3 = 8 or, x = 2

∴ the radius of the cone = 3 x 2 cm = 6 cm

∴ the diameter of the cone = 2 x 6 cm = 12 cm

Hence the diameter of the cone =12 cm.

Example 7. The radius of a right circular cone is 7 cm at its vertical angle is 60°. Find the curved surface area of the cone.

Solution:

Given:

The radius of a right circular cone is 7 cm at its vertical angle is 60°.

The semi-vertical angle is \(\frac{60^{\circ}}{2}\) = 30°.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Radius Of Right Circular Cone

 

Now, from the right-angled ΔAOB, we get,

\(\frac{\mathrm{AB}}{\mathrm{OB}}\) = cosec 30° or, \(\frac{\mathrm{AB}}{7}\) = 2

[∵ OB = radius = 7 cm and cosec 30° = 2]

or, AB = 14

∴ the height of the cone = 14 cm.

∴ the curved surface area of the cone = \(\frac{22}{7}\) x 7 x 14sq-cm = 308 sq-cm.

Hence the curved surface area of the cone = 308 sq-cm.

Example 8. If the height and the area of base of a right circular cone be increased by 4 times, then how many times the volume of the cone will be increased?

Solution:

Given:

If the height and the area of base of a right circular cone be increased by 4 times

Let the initial radius of the base and height of the cone be r unit arid h unit respectively.

Now, if the area of the base be increased by 4 times, let the radius of the base will be r1 unit.

∴ \(\pi r_1^2=4 \pi r^2\)

⇒ \(r_1^2=4 r^2\)

⇒ \(r_1^2=(2 r)^2\)

⇒ r1 = 2r

If the initial volume of the cone be V cubic-unit, and the volume after the increment be V1 cubic-unit, then

V = \(\frac{1}{3}\)πr2h and V1 = \(\frac{1}{3}\)π .(2r)2 .4h = 16.\(\frac{1}{3}\)πr2h = 16V. [∵ V = \(\frac{1}{3}\)πr2h]

Hence the volume of the cone will be 1 6 times of its initial volume, if its height and area of base is increased by 4 times.

Example 9. The curved surface area of a right circular cone is 42 times of the area of the base. Then what is the vertical angle of the cone?

Solution:

Given:

The curved surface area of a right circular cone is 42 times of the area of the base.

Let the radius of the base of the right circular cone be r unit and its slant height be l unit.

As per question, πrl = √2πr2 ⇒ l = √2r

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Curved Surface Area of A Right Circular Cone

 

Now, let the semi-vertical angle be θ.

∴ from the right-angled ΔAOB we get, \(\frac{OB}{AB}\)= sinθ

⇒  sinθ = \(\frac{r}{\sqrt{2} r}=\frac{1}{\sqrt{2}}\) = sin45°

⇒ θ = 45°.

Hence the vertical angle = 2 x 45° = 90°.

Example 10. The ared of bases of two right circular cones are equal. If the ratio of the slant heights be 2 : 3, then what will be the ratio of their curved surface areas?

Solution:

Given:

The ared of bases of two right circular cones are equal. If the ratio of the slant heights be 2 : 3

The area of the bases of the right circular cones are equal.

The radii of the bases of two cones will be also be equal.

Let the equal radii of the two cones be r unit and their slant heights be 21 units and 3l units respectively.

So, the ratio of their curved surface areas = πr2l: πr3l = 2:3.

Hence the required ratio of their curved surface areas is 2 : 3.

Class 10 Maths Solutions Wbbse

Mensuration Chapter 4 Right Circular Cone Long Answer Type Questions

Example 1. If the height and slant height of a cone are 6 cm and 10 cm respectively. Then determine the total surface area and the volume of the cone.

Solution:

Given:

If the height and slant height of a cone are 6 cm and 10 cm respectively

The height of the right circular cone is 6 cm and its slant height is 10 cm.

Let the radius of the cone be r cm.

∴ 62 + r2 = 102 ⇒ r2 = 100 – 36 = 64 ⇒ r = √64=8

∴ the radius of the cone is 8 cm.

∴ the total surface area of the cone = π x 8 (8 + 10) sq-cm = \(\frac{22}{7}\) x 8 x 18 sq-cm = 144π sq-cm.

Again, the volume of the cone = \(\frac{1}{3}\)π x (8)2 x 6 cubic- cm = 128π cubic- cm.

Hence the total surface area of the cone = 144 n sq-cm and volume = 128 7t cubic-cm.

Example 2. 77 sq-cm tripal is required to make a right circular conical tent. If the slant height of the tent is 7 m, then calculate the base area of the tent.

Solution:

Given:

77 sq-cm tripal is required to make a right circular conical tent. If the slant height of the tent is 7 m,

The slant height of the tent is 7 metres.

Let the radius of the base of the tent be r metres.

∴ the curved surface area of the tent = \(\frac{22}{7}\) x r x 7 sq-m = 22r sq-m.

As per question, 22r = 77 ⇒ r = \(\frac{77}{22}=\frac{7}{2}\)

∴ the area of the base of the tent = \(\frac{22}{7} \times\left(\frac{7}{2}\right)^2\) sq-m = \(\frac{22}{7} \times \frac{7}{2} \times \frac{7}{2}\) sq-m

= \(\frac{77}{2}\) sq- m = 38.5 sq- m.

∴ the required base area of the tent = 38.5 sq-m.

Example 3. The length of the base diameter of a wooden toy of conical shape is 10 cm. The expenditure for polishing whole surfaces of the toy at the rate of 2.10 per square-metre is 429. Calculate the height of the toy. Also determine the quantity of wood which is required to make the toy.

Solution:

Given:

The length of the base diameter of a wooden toy of conical shape is 10 cm. The expenditure for polishing whole surfaces of the toy at the rate of 2.10 per square-metre is 429.

The base diameter of the toy = 10 cm.

∴ Radius of the base of the toy = \(\frac{10}{2}\) cm = 5cm.

Let the slant height of the toy = l cm

∴ the curved surface area of the toy = \(\frac{22}{7}\) x 5 x / sq-cm

₹2.10 is expent to polish 1 sq-m of the area.

∴ ₹1 is expent to polish \(\frac{1}{2 \cdot 10}\) sq-m of the area.

∴ ₹429 is expent to polish \(\frac{1 \times 429}{2 \cdot 10}\) sq-m of the area.

As per condition, \(\frac{22}{7}\) x 5 x l = \(\frac{429}{2 \cdot 10}\)

⇒ l = \(=\frac{429 \times 7}{2 \cdot 10 \times 22 \times 5}\) ⇒ l = 13

So, the slant height of the toy = 13 cm.

Let the height of the toy be h cm

∴ h2 + 52 = (13)2 or, h2 + 25 = 169 or, h2 = 169 – 25 or, h2 = 144

∴ h = √144 ⇒ h = 12

∴ the height of the toy = 12 cm

Again, the volume of the toy = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 52 x 12 cc = \(\frac{2200}{7}\)cc = \(314 \frac{2}{7}\)cc

Hence the height of the toy is 12 cm and the quantity of wood required to make the toy is 314\(\frac{2}{7}\)cc.

Example 4. The quantity of iron-sheet to make a boya of right circular conical shape is 75\(\frac{3}{7}\)sq-m. If the slant height of it is 5 in, then calculate the volume of air in the boya and its height. Determine of the expenditure to colour the whole surface of the boya at the rate of ₹2.80 per square metres. [The width of the iron sheet not to be considered while calculating.]

Solution:

Given:

The quantity of iron-sheet to make a boya of right circular conical shape is 75\(\frac{3}{7}\)sq-m. If the slant height of it is 5 in, then calculate the volume of air in the boya and its height.

The slant height of the boya is 5 metres.

Let the radius, of the boya be r metres.

∴ the total surface area of the boya = \(\frac{22}{7}\)r (5+r) sq-m

∴ \(\frac{22}{7}\)r (5+r) = 75\(\frac{3}{7}\)

or, \(\frac{22}{7}\)r(5 + r) = \(\frac{528}{7}\)

or, r (r+5) = \(\frac{528 \times 7}{7 \times 22}\) or, r(r+5)= 24

or, r2 + 5r- 24 = 0 or, r2 + 8r- 3r- 24 = 0 or, r (r + 8) – 3 (r + 8) = 0 or, (r + 8) (r – 3) = 0

∴ either r + 8 = 0 or, r – 3 = 0

⇒ r = -8  ⇒ r = 3

Since the radius can never be negative, ∴ r = 3.

Now, if the height of the boya be h m, then h2 + 32 = 52 or, h2 = 25-9 or, h2 = 16 or, h = 4.

∴ the height of the boya is 4 metres.

∴ the volume of boya = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 32 x 4 cubic-metres

= \(\frac{264}{7}\) cubic-metres = 37\(\frac{5}{7}\) cubic-metres

Hence the height of the boya = 4 metres and the volume of air in. the boya =37\(\frac{5}{7}\) cubic- metres.

Also, at the rate of ₹2.80 per square ‘metre, the cost of colouring the whole surface of the boya

= ₹2.80 x \(\frac{528}{7}\) = ₹211.20

Class 10 Maths Solutions Wbbse

Example 5. In a right circular conical tent 11 persons can stay, For each person 4 sq-m space in the base and 20 m3 air are necessary. Determine the height of the tent put up exactly for 11 persons.

Solution:

Given:

In a right circular conical tent 11 persons can stay, For each person 4 sq-m space in the base and 20 m3 air are necessary.

Let the radius of the base of the tent be r metre.

∴ the area of the base = πr2 sq-metre

Again, space required for staying for 11 persons =11×4 sq-m. = 44 sq-m.

∴ \(\frac{22}{7}\) x r2 =44 ⇒ r2 =14.

Again, the total air required for 11 person = 20 x 11 m3 = 220 m3.

Now, if the height of the.terit be h metre, then its volume = \(\frac{1}{3}\)πr2h cubic- metres.

As per question, \(\frac{1}{3}\) x \(\frac{22}{7}\) x r2h = 220.

or, \(\frac{1}{3}\) x \(\frac{22}{7}\) x 14 x h = 220 [∵ r2 = 14]

or, h = \(\frac{220 \times 3 \times 7}{22 \times 14}\) or, h = 15

Hence the height of the tent = 15 metres.

Example 6. The external diameter of a conical-coronet made off thermocol is 21 cm in length. To wrap up the outer surface of the coronet with foil, the expenditure will be ₹57.75 at the rate of 10p per-m2. Calculate the height and slant height of the coronet.

Solution:

Given:

The external diameter of a conical-coronet made off thermocol is 21 cm in length. To wrap up the outer surface of the coronet with foil, the expenditure will be ₹57.75 at the rate of 10p per-m2.

The external diameter of the coronet = 21 cm.

∴ the external diameter of the coronet = \(\frac{21}{2}\)

Let the slant height of the coronet be 1 cm,

∴ the curved surface area = \(\frac{22}{7}\) x \(\frac{21}{2}\) x l sq-cm = 33 l sq-cm

10 paise = ₹0.1 is spent to wrap up 1 sq-cm

∴ ₹57.75 is spent to wrap up \(\frac{1 \times 57.75}{0 \cdot 1}\) sq -cm.= 577.5 sq-cm.

As per question, 33 l = 577.5 ⇒ l = \(\frac{577 \cdot 5}{33}\) = 17.5

Let the height of the coronet be h cm.

∴ h2 + \(\left(\frac{21}{2}\right)^2\) = (17.5)2 or, h2 + 110.25 = 306.25 or, h2 = 306.25 – 110.25

or, h2 = 196 or, h = 14.

Hence the height of the coronet is 14 cm and the slant height is 17.5 cm.

Example 7. The base area of a right circular cone is 21 m and height is 14 m. Calculate the expenditure to colour the curved surface at the rate of ₹1.50 per sq-m.

Solution:

Given:

The base area of a right circular cone is 21 m and height is 14 m.

The diameter of the base of the cone = 21 metres

∴ the radius of the base of the cone = \(\frac{21}{2}\) metres

The height of the cone = 14 metres

∴ the slant height of the cone = \(\sqrt{(14)^2+\left(\frac{21}{2}\right)^2}\) metres = \(\sqrt{196+\frac{441}{4}}\) metres

= \(\sqrt{\frac{784+441}{4}}\) metres = \(\sqrt{\frac{1225}{4}}\) metres = \(\frac{35}{2}\) metres.

∴ the curved surface area = \(\frac{22}{7} \times \frac{21}{2} \times \frac{35}{2}\)sq- metres = 577.5 sq-m.

So, the expenditure to colour the curved surface area at the rate of ₹1.50 per sq-m = ₹577.5 x 1.50 = ₹866.25

Hence the required cost = ₹866.25

Class 10 Maths Solutions Wbbse

Example 8. The shape of a heap of wheat is a right circular cone, the diameter of base of which is 9 metres and height is 3.5 metres. Find the volume of the heap of wheat. How much sq-metre of plastic sheet will be needed to cover that heap of wheat? [Given that n = 3.14 and 130 = 11.4]

Solution:

Given:

The shape of a heap of wheat is a right circular cone, the diameter of base of which is 9 metres and height is 3.5 metres.

The radius of the heap of wheat = \( metres

Height = 3.5 metres.

∴ the volume of the heap of wheat = [latex]\frac{1}{3}\) x π x \(\left(\frac{9}{2}\right)^2\) x 3.5 cubic- metres

= \(\frac{1}{3}\) x 3.14 x (4.5)2 x 3.5 cubic-metres = 74.18 cubic-metres

If the slant height of the heap of wheat be l metre, then

\(l^2=(3 \cdot 5)^2+\left(\frac{9}{2}\right)^2=\left(\frac{7}{2}\right)^2+\left(\frac{9}{2}\right)^2=\frac{49}{4}+\frac{81}{4}=\frac{130}{4}\)

∴ l = \(\sqrt{\frac{130}{4}}\) = 5.7 (approx.)

∴the curved surface area of the heap of wheat

= π x \(\frac{9}{2}\) x 5.7sq-m= 3.14 x 4.5 x 5.7 sq-m = 80.54 sq-m.

Hence the total volume of the heap of wheat is 7418 cubic metres and to cover the heap 80.54 sq-m of the plastic sheet will be needed.

Example 9. What will be the height of the cone of radius 12 cm if a solid cone is made by melting a solid cone of radius of base 6 cm and of slant height 10 cm?

Solution: The radius of the base of the first right circular cone = 6 cm and slant height = 10 cm.

∴ the height of the right circular cone = \(\sqrt{(10)^2-(6)^2}\) cm = 8cm

∴ volume of the cone = \(\frac{1}{3}\)π x 62 x 8 cu-cm

The new radius of the cone is 12 cm and the height is h cm, the volume of the cone = \(\frac{1}{3}\)π x (12)2 x h cubic- cm.

As per question, \(\frac{1}{3}\)π x(12)2 xh = \(\frac{1}{3}\)π x (6)2 x 8

⇒ h = \(\frac{6 \times 6 \times 8}{12 \times 12}\) = 2.

Hence the height of the new cone = 2 cm.

Example 10. What changes will be occured in the volume of a right circular cone if its radius be increased by 10% and its height be decreased by 10%?

Solution: Let the radius of the right circular cone be r units and height be h units.

∴ the volume of the cone = \(\frac{1}{3}\)πr2h cubic-units

Now, if the radius be Increased by 10%, then the radius become (r + r x \(\frac{10}{100}\))units = \(\frac{11r}{10}\)units

Again, if the height of the cone be decreased by 10%, then its height becomes

\(\left(h-h \times \frac{10}{100}\right)\) units = \(\frac{9h}{10}\)units.

As a result of these increasing and decreasing, the volume of the cone becomes

\(\pi \times\left(\frac{11 r}{10}\right)^2 \times \frac{9 h}{10}\)cu – units = \(\pi \times \frac{121 r^2}{100} \times \frac{9 h}{10} \)cubic-units = \(\frac{1089}{1000} \pi r^2 h\)cubic -units

 

∴ the increment in volume = \(=\left(\frac{1089}{1000} \pi r^2 h-\frac{1}{3} \cdot \pi r^2 h\right)\)cu-units = \(\frac{2267}{3000}\)πr2h cubic -units.

So, the percentage of increment in volume = \(=\frac{\frac{2267}{3000} \pi r^2 h}{\pi r^2 h} \times 100 \%=\frac{2267}{3000} \times 100 \%\) = 75.57%

Hence the volume of the cone will be increased by 75.57 % (approx.)

Example 11. If the whole surface area of a right circular cone be 3696 sq-cm and the ratio of its radius of base and height be 3: 4, then what will be the curved surface area of the cone?

Solution:

Given:

If the whole surface area of a right circular cone be 3696 sq-cm and the ratio of its radius of base and height be 3: 4,

The ratio of the radius of base and height of the cone is 3: 4; then let the radius of the cone be 3x cm and height be 4x cm.

So, the slant height of the. cone = \(\sqrt{(3 x)^2+(4 x)^2}\) cm = \(\sqrt{25 x^2}\)cm = 5x cm.

The whole surface area of the cone = 3696 sq-cm

∴ \(\frac{22}{7}\) x 3x x(3x + 5x) = 3696

or, \(\frac{22}{7}\) x 3x x 8x = 3696

or, \(x^2=\frac{3696 \times 7}{22 \times 3 \times 8}\)

or, x2 = 49 or, x = √49 = 7.

So, the radius of base of the cone = 3x cm = 3 x 7 cm = 21 cm

and slant height of the cone = 5x cm = 5 x 7 cm = 35 cm

∴ Curved surface area of the cone = \(\frac{22}{7}\) x 21 x 35 sq-cm = 2310 sq-cm

Hence the required curved surface of the right circular cone = 2310 sq-cm.

Class 10 Maths Solutions Wbbse

Example 12. If the curved surface area of a toy of shaped a right circular cone of height 24 cm is 550 sq-cm, then what will be the volume of the toy?

Solution:

Given:

If the curved surface area of a toy of shaped a right circular cone of height 24 cm is 550 sq-cm

Let the radius of base of the toy be r cm and its slant height be l cm.

Then, l2 = r2 + 242 [∵ height = 24 cm] ⇒ l2 = r2 + 576 ⇒ l = \(\sqrt{r^2+576}\).

The curved surface area of the toy = 550 sq-cm.

∴ \(\frac{22}{7}\) x r x l = 550

or, r x \(\sqrt{r^2+576}\) = \(\frac{550 \times 7}{22}\)

or, r x \(\sqrt{r^2+576}\) = 25 x 7

or, r2 x (r2 + 576) = (25 x 7)2 or, r4 + 576 r2 = 625 x 49 or, r4 + 576r2 – 625 x 49 = 0

or, r4 + (625 – 49)r2 – 625 x 49 = 0 or, r4 + 625r2 – 49 r2 – 625 x 49 = 0

or, r2 (r2 + 625) – 49 (r2 + 625) = 0 or, (r2 + 625) (r2 – 49) = 0

∴ either r2 + 625 = 0, which is impossible, or, r2 – 49 = 0 ⇒ r2 = 72 ⇒ r = 7.

So, the radius of the base of the toy = 7

∴ the volume of the toy = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 72 x 24 cubic-cm

= \(\frac{1}{3}\) x \(\frac{22}{7}\) x 49 x 24 cc = 1232 cc

Hence the required volume of the toy = 1232 cc.

Example 13. The volume of a right circular cone is 410\(\frac{2}{3}\)cc. If the radius of the base of the cone be 14 cm, then what will be the curved surface area of the cone?

Solution:

Given:

The volume of a right circular cone is 410\(\frac{2}{3}\)cc. If the radius of the base of the cone be 14 cm

Let the height of the cone be h cm.

The radius of the cone = 14 cm,

∴ the volume of the cone = \(\frac{1}{3}\) x \(\frac{22}{7}\) x 142 x h cc.

As per question, \(\frac{1}{3}\) x \(\frac{22}{7}\) x 142 x h = 410\(\frac{2}{3}\)

or, \(\frac{22}{3}\) x 28 h = \(\frac{1232}{3}\)

or, h = \(\frac{1232 \times 3}{3 \times 22 \times 28}\) or, h = 2

∴ the height of the cone = 2 cm.

∴ the slant height of the cone = \(\sqrt{(14)^2+(2)^2}\)cm = 200 cm = \(\sqrt{100 \times 2}\) cm = 10√2 cm

∴ the curved surface area of-the cone = \(\frac{22}{7}\) x 14 x 10V2 sq- cm = 440√2 sq- cm

Hence the required curved surface of the cone = 440√2 sq-cm

Example 14. What changes will be occured in the height of a right circular cone if its radius of base is decreased by 50% and its volume be decreased by 25%?

Solution: Let the radius of base of the cone be r units and height be h units.

Now, if the volume of the cone be V cubic-units, then V = \(\frac{1}{3}\)πr2h…..(1)

Now, let the volume of the cone be V1 cubic-unit and height be h1 units when its radius of base is decreased by 50% and volume is decreased by 25%.

∴ \(\quad \mathrm{V}_1=\frac{1}{3} \pi\left(r-r \times \frac{50}{100}\right)^2 \times h_1=\frac{1}{3} \pi\left(r-\frac{r}{2}\right)^2 \times h_1=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

i.e, \(\mathrm{V}-\mathrm{V} \times \frac{25}{100}=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

or, \(\frac{3 \mathrm{~V}}{4}=\frac{1}{3} \pi \times \frac{r^2}{4} \times h_1\)

or, \(9 \mathrm{~V}=\pi r^2 h_1\)

or, \(9 \times \frac{1}{3} \pi r^2 \dot{h}=\pi r^2 h_1\) [by (1)]

or, \(3 h=h_1\)

∴h1=3h

∴ the percentage of change in height = \(\frac{3 h-h}{h}\) x 100% = \(\frac{2h}{h}\) x 100% =200%

Hence the required increase in height in percent of the cone = 200%.

Example 15. What changes in curved surface area of a right circular cone will be occured if its radius of base and height is doubled?

Solution:

Let the radius of base, height and slant height of the right circular cone be r unit, h unit and l unit respectively.

∴ l = \(\sqrt{h^2+r^2}\)…….(1)

Now if the curved surface area of the cone be A sq-unit, then

A = πrl……(2)

Also, if the radius of base and the height of the cone be doubled, then let the slant height of it be l1 units

∴ \(l_1=\sqrt{(2 r)^2+(2 h)^2}=\sqrt{4 r^2+4 h^2}=\sqrt{4\left(r^2+h^2\right)}=2 \sqrt{h^2+r^2}\) = 2l [by (1)]…..(3)

In this case, if the curved surface area be A1 sq-unit, then

A1 = π x 2r x lπ = π x 2r x 2l [by (3)] = 4πrl = 4A [by (2)]

Hence if the radius of base and height of a right circular cone be doubled, then the curved surface area becomes 4 times of its previous curved surface area.

Class 10 Maths Solutions Wbbse

Example 16. If the curved surface area, volume, height and semi-vertical angle of a right circular cone be S,V.h and respectively, then prove that S= \(\frac{\pi h^2 \sin a}{\cos ^2 a}\) and V = \(\frac{1}{3} \pi h^3 \tan ^2 \alpha\)

Solution:

Given:

If the curved surface area, volume, height and semi-vertical angle of a right circular cone be S,V.h and respectively,

Here,, the height, AB = h, ∠BAC = α = semi-vertical angle, radius of base, BC = r and slant height, AC = l.

∴ \(\frac{r}{h}\) = tan α or, h tan α= r……(1)

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Semi Verticle Of A Right Circular Cone

 

and \(\frac{h}{l}\) = cosα or, l = \(\frac{h}{\cos \alpha}\)……(2)

Now, the curved surface are of the cone = πrl.

= π x h tan α x \(\frac{h}{\cos \alpha}\) [by (1) and (2)]

= x h.\(\frac{\sin x}{\cos \alpha} \times \frac{h}{\cos \alpha}=\frac{\pi h^2 \sin \alpha}{\cos ^2 \alpha}\)

∴ S = \(\frac{\pi h^2 \sin \alpha}{\cos ^2 \alpha}\)

Moreover, the volume of the cone, V = \(\frac{1}{3}\)πr2h = \(\frac{1}{3}\) (h tan α)2.h [by (1)]

= \(\frac{1}{3} \pi \cdot h^2 \tan ^2 \alpha \cdot h=\frac{1}{3} \pi h^3 \tan ^2 \alpha\)

Hence V = \(\frac{1}{3} \pi \cdot h^3 \tan ^2 \alpha\)

Example 17. If the height, curved surface area and volume of a right circular cone be A, c and v respectively, then prove that 3πvh3 – c2h + 9v2 = 0.

Solution:

Given:

If the height, curved surface area and volume of a right circular cone be A, c and v respectively

Let the radius of base and slant height of the cone be r unit and l unit respectively.

∴ l2 = h2 + r2……(1)

Again, v = \(\frac{1}{3}\)πr2h ……(2) and c = πrl…….(3)

∴ 3πvh3 – c2h2 + 9V2

= 3π x \(\frac{1}{3}\)πr2h x h3 -(πrl)2 x h2 +9 x (\(\frac{1}{3}\)πr2h)2 [by (2) and (3)]

= \(\pi^2 r^2 h^4-\pi^2 r^2 l^2 h^2+\pi^2 r^4 h^2\)

= \(\pi^2 r^2 h^4-\pi^2 r^2\left(r^2+h^2\right) \cdot h^2+\pi^2 r^4 h^2\) [by (1)]

= \(\pi^2 r^2 h^4-\pi^2 r^4 h^2-\pi^2 r^2 h^4+\pi^2 r^4 h^2\) = 0 .

Hence 3πvh3 – c2h2 + 9v2 = 0 [Proved]

Example 18. The height of a right circular cone is 30 cm. A small part of the cone is cut off from the above part of the cone with the help of a plane parallel to the base of the cone. If the volume of the small part be 1/27 part of the volume of the original cone, then at what distance above the base of the cone, it has been cut off?

Solution:

Given:

The height of a right circular cone is 30 cm. A small part of the cone is cut off from the above part of the cone with the help of a plane parallel to the base of the cone. If the volume of the small part be 1/27 part of the volume of the original cone

Let OAB is the original cone of radius of base R cm and its height be H cm.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Heigth Of The Right Circular Cone

 

∴ the volume of the cone = \(\frac{1}{3}\)πR2Hcc = \(\frac{1}{3}\)πR2 X 30CC = 10πR2cc

The cut-off small part of the cone is OCD, the radius of base of which is r cm and height is h cm.

Then the volume of this part = \(\frac{1}{3}\)πr2h cc.

As per question, \(\frac{1}{3}\)πr2h = \(\frac{1}{27}\)(10πR2)

⇒ h = \(\frac{10 \pi \mathrm{R}^2}{27} \times \frac{3}{\pi r^2}\)

h= \(\frac{10}{9} \cdot \frac{\mathrm{R}^2}{r^2}\)…….(1)

Now, from ΔOQB and ΔOPD we get, \(\frac{\mathrm{QB}}{\mathrm{PD}}=\frac{\mathrm{OQ}}{\mathrm{OP}}=\frac{30}{h} \Rightarrow \frac{\mathrm{R}}{r}=\frac{30}{h}\)

∴ from (1) we get, h = \(\frac{10}{9} \cdot\left(\frac{\mathrm{R}}{r}\right)^2=\frac{10}{9} \cdot\left(\frac{30}{h}\right)^2=\frac{10}{9} \times \frac{900}{h^2}\)

or, h3 = 1000 or h3 = (10)3 or, h = 10

∴ the small part is cut off at a distance of (30- 10) cm = 20 cm high above the base of the cone.

Hence the required height = 20 cm.

Class 10 Maths Solutions Wbbse

Example 19. The lengths of the adjacent sides of right angle of a right-angled triangle are 20 cm and 15 cm. What will be the total volume of the two right circular cones when this triangle is rotated with its hypotenuse taken as the axis?

Solution:

Given:

The lengths of the adjacent sides of right angle of a right-angled triangle are 20 cm and 15 cm.

WBBSE Solutions For Class 10 Maths Mensuration Chapter 4 Right Circular Cone Lengths of the Adjacent Sides Of Right Angle

 

Let the perpendicular of the right-angled triangle ABC be ABC = 20 cm and its base, AB = 15 cm.

∴ hypotenuse, AC = \(\sqrt{(20)^2+(15)^2} \mathrm{~cm}\)

= \(\sqrt{400+225} \mathrm{~cm}=\sqrt{625} \mathrm{~cm}=25 \mathrm{~cm}\)

Now, perpendicular BO is drawn from B to the hypotenuse AC.

∴ the area of the ΔABC = \(\frac{1}{2}\) x AB x BC = \(\frac{1}{2}\) x AC x BO

⇒ AC x BO = AB x BC ⇒ 25 x BO = 15 x 20

⇒ BO = \(\frac{15 \times 20}{25}\) => BO = 12

So, the radius of two cones produced by rotating the triangle with the hypotenuse AC as the axis will be 12 cm.

Also, the heights of the two cones will be AO and CO.

So, the total volume of the two cones = \(\left(\frac{1}{3} \pi \times 12^2 \times \mathrm{AO}+\frac{1}{3} \pi \times 12^2 \times \mathrm{CO}\right)\)cc

= \(\frac{1}{3}\)π x (12)2 x (AO + BO)cc = \(\frac{1}{3}\)π x 144 x AC cc

= \(\frac{1}{3}\) x 144 x 25π cc = 1200π cc

Hence the total volume of the two produced cones = 1200π cc.

 

 

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 3 Sphere

Mensuration Chapter 3 Sphere

Sphere 

If a semi-circle is rotated through an angle of 360° with its diameter as the axis, then the solid object thus produced is called a sphere.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 3 Sphere Semi circle Is Rotated Through An Angle Of 360 Degree

For example, a football is a sphere. We see various spherical objects in our daily life, such as marbles ball, bullets, etc.

Surface of a sphere:

It is obvious that the surface of a sphere is a curved surface and the number of the curved surfaces is one.

So, we can say that the solid object surrounded by only one curved surface is known as a sphere. There is no plane surface in a sphere.

WBBSE Solutions for Class 10 Maths

Centre of a sphere:

The centre of the semi-circle, the diameter of which was taken as the axis to construct the sphere is known as the centre of the sphere.

 

WBBSE Solutions For Class 10 Maths Mensuration Chapter 3 Sphere Centre Of The Sphere

 

O is the centre of the sphere.

The radius of a sphere:

The radius of the semi-circle, by the rotation of which the sphere is produced, is known as the radius of the sphere, i.e., the distance of any point on the circumference of the sphere from its centre is known as its radius.

OC is one of the radii of the sphere. Obviously, there can be infinite number of radii of a sphere, though the lengths of all of them are always equal.

The radius of the semi-circle and the radius of the sphere are the same.

AB is the diameter of the sphere.

Maths Solutions Class 10 Wbbse

Height of a sphere:

The length of any diameter of the sphere is called the height of the sphere.

AB is a diameter of the semi-circle and so it is also the height of the sphere.

Wbbse Class 10 Maths Solutions

Formulae related to sphere

Let the radius of a sphere be r units, then

  1. The curved surface or the total surface area of the sphere = 4nr2 sq-units.
  2. Height of the sphere = 2r units.
  3. Volume of the sphere = \(\frac{4}{3}\) πr3 cubic-units.

Formulae related to hollow spheres:

Let the external radius of a hollow sphere be R units and its internal radius be r units. Then

  1. the volume of the hollow sphere = \(\frac{4}{3}\)π(R3 -r3) cubic units.
  2. the area of outer curved surface = 4πR2 sq-units.
  3. the area of inner curved surface = 4πr2sq-units.

Formulae related to solid hemispheres:

Let the radius of the solid hemisphere be r units. Then

  1. the area of curved surface of the solid hemisphere = 2πr2 sq-units.
  2. the area of plane surface of the solid hemisphere = πr2 sq-units.
  3. the total area of the surfaces of the solid hemisphere = (2πr2 + πr2) sq-units = 3πr2 sq-units
  4. the volume of the solid hemisphere = \(\frac{2}{3}\)πr3 cubic-units

Formulae related to a hollow hemisphere:

Let the radius of the hollow hemisphere be r units, and then

  1. Whole surface area = curved surface area = 2πr2 sq-units
  2. Volume of the hollow hemisphere = \(\frac{2}{3}\) πr3 cubic-units.
  3. Area of the whole surface of a hollow hemisphere = {2πR2 + 27πr2 + (πR2 – πr2)} sq-units, where R and r are the radii of the outer and inner radius of the hollow hemisphere.

 

Mensuration Chapter 3 Sphere Multiple Choice Questions

Example 1. The volume of a solid sphere of radius 2r units is

  1. \(\frac{32 \pi r^3}{3}\) cubic-units
  2. \(\frac{16 \pi r^3}{3}\) cubic-units
  3. \(\frac{8 \pi r^3}{3}\) cubic-units
  4. \(\frac{64 \pi r^3}{3}\) cubic-units

Solution: Radius of the solid sphere = 2r units

∴ Volume of the solid sphere = \(\frac{4}{3}\)π x (2r)3 cubic-units

= \(\frac{2}{3}\) π x 8r3 cubic – units = \(\frac{32 \pi r^3}{3}\) cubic-units

∴ 1. \(\frac{32 \pi r^3}{3}\) cubic-units is correct

The volume of a solid sphere of radius 2r units is  \(\frac{32 \pi r^3}{3}\) cubic-units

Wbbse Class 10 Maths Solutions

Example 2. If the ratio of volumes of two solid spheres is 1: 8, the ratio of their curved surface areas is

  1. 1: 2
  2. 1: 4
  3. 1: 8
  4. 1: 16

Solution: Let the radii of the two spheres be r1 and r2 units.

So the volumes of two spheres are \(\frac{4}{3} \pi r_1^3\) cubic-units and \(\frac{4}{3} \pi r_2^3\) cubic-units.

As per question, \(\frac{4}{3} \pi r_1^3\) : \(\frac{4}{3} \pi r_2^3\) = 1:8

\(\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}=\frac{1}{8}\)

 

or, \(\frac{r_1^3}{r_2^3}=\frac{1}{8}\)

or, \(\left(\frac{r_1}{r_2}\right)^3=\left(\frac{1}{2}\right)^3\)

or, \(\frac{r_1}{r_2}=\frac{1}{2}\)

= \(\frac{4}{3} \pi r_1^3\) : \(\frac{4}{3} \pi r_2^3\)

= \(\frac{4 \pi r_1^2}{4 \pi r_2^2}=\left(\frac{r_1}{r_2}\right)^2=\left(\frac{1}{2}\right)^2\) = 1:4

Hence 2. 1: 4 is correct

The ratio of their curved surface areas is 1: 4

Maths Solutions Class 10 Wbbse

Example 3. The whole surface area of a solid hemisphere with length of 7 cm radius is

  1. 588π sq-cm
  2. 392π sq-cm
  3. 147π sq-cm
  4. 98πsq-cm

Solution: The radius of the solid hemisphere = 7 cm.

∴ the whole surface area of it = 3π x (7)2 sq-cm = 3π x 49 sq-cm = 147π sq-cm.

∴ 3. 147π sq-cm is correct.

The whole surface area of it 147π sq-cm

Example 4. If the ratio of curved surface areas of two solid spheres is 16 : 9, the ratio of their volumes is

  1. 64: 2 7
  2. 4: 3
  3. 27: 64
  4. 3: 4

Solution: Let the radii of two solid spheres be units r1 and r2 units respectively.

∴ the curved surface areas of the spheres are \(4 \pi r_1^2\) sq-units and \(4 \pi r_2^2\) sq-units respectively.

As per question, \(4 \pi r_1^2\) : \(4 \pi r_2^2 s q\)

⇒ \(\frac{4 \pi r_1^2}{4 \pi r_2^2}=\frac{16}{9}\)

⇒ \(\left(\frac{r_1}{r_2}\right)^2=\left(\frac{4}{3}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{4}{3}\)

∴ The ratio of their volumes

= \(\frac{4}{3} \pi r_1^3:=\frac{4}{3} \pi r_2^3\)

= \(\frac{\frac{4}{3} \pi r_1^3}{\frac{4}{3} \pi r_2^3}\)

= \(\left(\frac{r_1}{r_2}\right)^3=\left(\frac{4}{3}\right)^3=\frac{64}{27}\)

= 64: 27

∴ 1. 64: 27 is correct

The ratio of their volumes is 64: 27

Maths Solutions Class 10 Wbbse

Example 5. If the numerical value of curved surface area of a solid sphere is three times of its volume, the length of its radius is

  1. 1 unit
  2. 2 units
  3. 3 units
  4. 4 units

Solution: Let the radius of the sphere = r units

∴ curved surface area of the sphere = 4 πr2 sq-units

∴ the volume of the sphere = \(\frac{4}{3}\)πr3 cubic-units

As per questions, 4πr2 = 3x \(\frac{4}{3}\) πr3

or, 4πr2 = 4πr3

⇒ 1 = r ⇒ r = 1

∴ radius = 1 unit

∴ 1. 1 unit is correct.

The length of its radius is 1 unit

Example 6. The radius of a copper sphere is twice the radius of an iron sphere. The numerical value of the whole surface area of the copper sphere is equal to the numerical value of the volume of the iron sphere. Then the radius of the copper sphere will be

  1. 6 units
  2. 12 units
  3. 18 units
  4. 24 units

Solution: Let the radius of the iron sphere be r units.

∴ the radius of the copper-sphere is 2r units.

∴ the whole surface area of the copper-sphere = 4π x (2r)2 sq-units = 16πr2 sq-units

Also, the volume of the iron sphere = \(\frac{4}{3}\)πr3 cubic-units

As per question, \(\frac{4}{3}\)πr3 = 16πr2 ⇒ r = 12

∴ the radius of the copper-sphere = 2 x 12 units = 24 units.

∴ 4. 24 units is correct.

The radius of the copper sphere will be 24 units

Example 7. If the volume of a sphere be V cubic units, then the radius of the sphere will be

  1. \(\left(\frac{3 V}{4 \pi}\right)^3\) units
  2. \(\frac{3 \mathrm{~V}}{4 \pi}\) units
  3. \(\sqrt{\frac{3 V}{4 \pi}}\) units
  4. \(\sqrt[3]{\frac{3 V}{4 \pi}}\) units

Solution: Let the radius of the sphere be r units

∴ the volume of the sphere = \(\frac{4}{3}\)πr3cubic-units

As per question, \(\frac{4}{3}\)πr3 = V

or, r3 = \(\frac{3 \mathrm{~V}}{4 \pi}\) or r = \(\sqrt[3]{\frac{3 V}{4 \pi}}\)

∴ 4. \(\sqrt[3]{\frac{3 V}{4 \pi}}\) is correct.

The radius of the sphere will be \(\sqrt[3]{\frac{3 V}{4 \pi}}\)

Example 8. If the radius of a sphere is decreased by \(\frac{1}{2}\) times times, then the curved surface area of the sphere will be changed by

  1. \(\frac{1}{2}\) times
  2. \(\frac{1}{4}\) times
  3. \(\frac{1}{8}\) times
  4. \(\frac{1}{16}\) times

Solution: If the radius of a sphere be r units, then the curved surface area of it will be = S = 4πr2 sq-units.

If the radius is decreased by \(\frac{1}{2}\) times times, then the curved surface area

= 4 x \(\left(\frac{r}{2}\right)^2\) sq – units = πr2 sq-units = \(\frac{1}{4}\) x 4πr2 sq-units

= \(\frac{1}{4}\) x S sq – units = \(\frac{S}{4}\) sq – units

∴ the new curved surface area of the sphere will be \(\frac{1}{4}\) times of the previous curved surface area.

∴ 2. \(\frac{1}{4}\) times is correct.

The curved surface area of the sphere will be changed by \(\frac{1}{4}\) times

Wbbse Class 10 Maths Solutions

Example 9. If the volume of a sphere is increased by 8 times, then its radius will be increased by

  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times

Solution: Let the radius of the sphere be r units and its volume be V cubic units.

∴ V = \(\frac{4}{3}\)πr3 (by formula)

Now, if the volume of the sphere be increased by 8 times, i.e., its volume be 8V cubic-units, then let its radius be r1, units.

∴ 8V = \(\frac{4}{3} \pi r_1^3\)

⇒ 8 x \(\frac{4}{3}\)πr3 = \(\frac{4}{3} \pi r_1^3\)

⇒ \(r_1^3=8 r^3\)

⇒ \(r_1^3=(2 r)^3\)

⇒ r1= 2r, i.e., the radius will be increased by 2 times.

∴ Hence 1. 2 times is correct.

Radius will be increased by 1. 2 times

Example 10. If the external and internal radii of a hollow sphere be 6 cm and 3 cm respectively, then the volume (i.e.. the material) of the sphere will be

  1. 629 cu-units
  2. 792 cu-units
  3. 829 cu-units
  4. 929 cu-units

Solution: The required volume of the sphere = \(\left(\frac{4}{3} \pi \times 6^3-\frac{4}{3} \pi \times 3^3\right)\)

= \(\frac{4}{3} \times \frac{22}{7}\)(216-27) cu.units

= \(\frac{4}{3} \times \frac{22}{7}\) x 189 cubic. units

= 792 cubic.units.

∴ 2. 792 cu-units is correct.

The volume (i.e.. the material) of the sphere will be 792 cu-units

 

Mensuration Chapter 3 Sphere True Or False

Example 1. If the ratio of curved surface areas of two hemispheres is 4: 9, then the ratio of their lengths of radii is 2 : 3

Solution: Let the radii of the two spheres be r1 and r2 units.

As per question, \(2 \pi r_1^2: 2 \pi r_2^2=4: 9\)

⇒  \(\frac{2 \pi r_1^2}{2 \pi r_2^2}=\frac{4}{9}\)

⇒ \(\left(\frac{r_1}{r_2}\right)^2=\left(\frac{2}{3}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{2}{3}\)

⇒ r1: r2 = 2: 3

Hence the given statement is true.

Wbbse Class 10 Maths Solutions

Example 2. If the length of radius of a solid sphere be doubled, the volume of sphere will be doubled.

Solution: Let the radius of the sphere be r units, then the volume of the sphere will be \(\frac{4}{3}\)πr3 cu-units.

Now, if the radius be doubled, the volume of the sphere will be \(\frac{4}{3}\) π x(2r)3 cu-units= 8 x \(\frac{4}{3}\)πr3 cubic- units.

i,e, the volume of the sphere will be 8 times of its previous volume.

Hence th given statement is false.

Mensuration Chapter 3 Sphere Fill In The Blanks

Example 1. The name of solid which is composed of only one surface is ______

Solution: sphere

Example 2. The number of surfaces of a solid hemisphere is ______

Solution: 1

Example 3. If the length of radius of a solid hemisphere is 2r units, its whole surface area is _______ πr2 sq-units.

Solution: 12

Mensuration Chapter 3 Sphere Short Answer Type Questions

Example 1. The numerical values of volume and whole surface area of a solid hemisphere are equal. Find the length of radius of the hemisphere.

Solution:

Given:

The numerical values of volume and whole surface area of a solid hemisphere are equal.

Let the radius of the solid hemisphere = r units.

∴ the volume of the solid hemisphere = \(\frac{2}{3}\)πr3 cu.units, and its whole surface area = 3πr2 sq.units.

As per question, \(\frac{2}{3}\)πr3 = 3πr2

⇒ r = \(\frac{9}{2}\) = 4.5

Hence the radius of the solid hemisphere = 4.5 units.

Example 2. The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of cylinder are 12 cm. Find the length of radius of the sphere.

Solution:

Given:

The curved surface area of a solid sphere is equal to the surface area of a solid right circular cylinder. The lengths of both height and diameter of cylinder are 12 cm.

Let the radius of the sphere be r cm.

∴ curved surface area of the solid sphere = 4πr2 sq-cm

The radius of the right circular cylinder = \(\frac{12}{2}\) cm= 6cm. and height of it = 12 cm.

∴ surface area of the right circular cylinder = 2π x 6 x 12 sq-cm.

As per question, 4πr2 = 2π x 6 x 12 or, r2 = 36 or, r = 6.

Hence the required radius of the sphere = 6cm.

Wbbse Class 10 Maths Solutions

Example 3. Whole surface area of a solid hemisphere is equal to the curved surface area of a solid sphere. Find the ratio of lengths of radius of hemisphere and sphere.

Solution:

Given:

Whole surface area of a solid hemisphere is equal to the curved surface area of a solid sphere.

Let the radius of the solid hemisphere be r1 unit and the radius of the solid sphere be r2 unit.

∴ the whole surface area of the solid hemisphere = \(3 \pi r_1^2\) sq-unit and the curved surface area of the solid sphere = \(4 \pi r_2^2\) sq-unit

As per question, \(3 \pi r_1^2\) = \(4 \pi r_2^2\)

⇒ \(\frac{r_1^2}{r_2^2}=\frac{4}{3}\)

⇒ \(\left(\frac{r}{r_2}\right)^2=\left(\frac{2}{\sqrt{3}}\right)^2\)

⇒ \(\frac{r_1}{r_2}=\frac{2}{\sqrt{3}}\) = 2: 3

Hence the required ratio = 2 : √3.

Example 4. If curved surface area of a solid sphere is S and volume is V, then find the value of \(\frac{S^3}{V^2}\) [putting the value of π].

Solution:

Given:

If curved surface area of a solid sphere is S and volume is V

Let the radius of the solid sphere be r unit.

Then the curved surface area of the sphere = 4πr2sq.units.

As per question, S = 4πr2 ……..(1)

Also, the volume of the sphere = \(\frac{4}{3}\)πr3cu.units.

As per question, V = \(\frac{4}{3}\)πr3 ………..(2)

∴ \(\frac{\mathrm{S}^3}{\mathrm{~V}^2}=\frac{\left(4 \pi r^2\right)^3}{\left(\frac{4}{3} \pi r^3\right)^2}\)

[Divding cube of (1) by square of (2)]

= \(\frac{64 \pi^3 r^6}{\frac{16}{9} \pi^2 r^6}=\frac{64 \times 9}{16} \pi=36 \pi\)

Hence the required value of \(\frac{\mathrm{S}^3}{\mathrm{~V}^2}\) = 36π.

Wbbse Class 10 Maths Solutions

Example 5. If the length of radius of a sphere is increased by 50%, how much percent will be increased of its curved surface area?

Solution:

Given:

If the length of radius of a sphere is increased by 50%,

Let the radius of the sphere be r unit.

∴ the area of the curved surface of the sphere = 4πr2sq-unit.

If the radius of the sphere be increased by 50% then the new radius will be \(\left(r+r \times \frac{50}{100}\right)\) unit = \(\frac{3r}{2}\) unit.

Then the curved surface area of the sphere will be \(4 \pi\left(\frac{3 r}{2}\right)^2\) sq.units = \(4 \pi \cdot \frac{9 r^2}{4}\) sq.units = 9nr1 sq.units.

∴ increase of curved surface area = (9πr2 – 4πr2) sq.unit = 5πr2sq.unit.

∴ the percent of increment of the curved surface area of the sphere = \(\frac{5 \pi r^2}{4 \pi r^2}\) x 100% = 125%

Hence the required percent = 125%

Example 6. The diameter of a sphere is double of the diameter of another sphere. Then how much times will be the volume of smaller sphere than the volume of the greater sphere?

Solution:

Given:

The diameter of a sphere is double of the diameter of another sphere.

Let the radius of the smaller sphere be r unit.

Then the radius of the greater sphere will be 2r unit, since the diameter of the greater sphere is double of the diameter of the smaller one.

Now, volume of the smaller sphere, V = \(\frac{4}{3}\)πr3cu.unit

and the volume of the greater sphere= \(\frac{4}{3}\)π x (2r)3cu. unit = \(\frac{4}{3}\)π x 8r3 cu.unit = 8 x \(\frac{4}{3}\) πr3cu.unit.

i.e., 8 times of the smaller sphere.

Hence the volume of the greater sphere is 8 times of the volume of the smaller sphere.

Example 7. What will be the ratio of diameter of a hemisphere and the length of its circumference?

Solution:

Let the radius of the hemisphere be r unit.

∴ its diameter = 2r unit

Also, the circumference of the hemisphere = (πr + 2r) unit.

As per question, 2r : (πr + 2r) = 2r : r (π + 2) = 2 : (π + 2)

Hence the ratio = 2 : (π + 2).

Example 8. If by melting a solid hemisphere be made into a sphere, then what will be the ratio of their radii?

Solution:

Let the radii of the hemisphere and the sphere be r1 unit and r2 unit respectively.

So, the volume of the hemisphere = \(\frac{2}{3} \pi r_1^3\) cubic.unit, and the volume of the original sphere = \(\frac{4}{3} \pi r_2^3\) cubic.unit

As per question, \(\frac{2}{3} \pi r_1^3=\frac{4}{3} \pi r_2^3\)

or, \(\frac{r_1^3}{r_2{ }^3}=\frac{4}{2}=2\)

or, \(\left(\frac{r_1}{r_2}\right)^3=2\)

or, \(\frac{r_1}{r_2}=\sqrt[3]{2}\)

∴ r1: r2 = \(\sqrt[3]{2}\) :1

Hence the required ratio = \(\sqrt[3]{2}\) : 1.

Example 9. A circular iron plate of thickness 2/3 cm is made by beating an iron sphere of diameter 4 cm. What will be the radius of the iron plate?

Solution:

Given:

A circular iron plate of thickness 2/3 cm is made by beating an iron sphere of diameter 4 cm.

The radius of the iron-sphere = \(\frac{4}{3}\) cm = 2cm.

∴ the volume of the iron-sphere = \(\frac{4}{3}\)π x 23 cc. = \(\frac{32 \pi}{3}\) cc.

Let the radius of the iron plate be r cm.

∴ the area of the iron-plate = πr2 sq-cm.

Since the iron-plate is of thickness \(\frac{2}{3}\) cm,its volume = \(\frac{2}{3}\)πr2 cu.cm.

As per question, \(\frac{2}{3}\)πr2 = \(\frac{32 \pi}{3}\)

⇒ r2 = 16 ⇒ r = √16 = 4

Hence the radius of the circular iron plate = 4 cm.

Example 10. The external radius of a hollow hemisphere is 6 cm and it is of thickness 2 cm. What will be the whole surface area of the hemisphere?

Solution:

Given:

The external radius of the hemisphere is 6 cm. and the thickness of it is 2cm.

∴ the internal radius of the hemisphere = (6 – 2) cm = 4 cm.

∴ the whole surface area of the hemisphere

= (2π x 62 + 2π x 42 + π x 62 – π x 42) sq.cm.

= π(2 x 36 + 2 x 16 + 36 – 16) sq.cm

= \(\frac{22}{7}\) x(72 + 32 + 20) sq.cm = \(\frac{22}{7}\) x 124 sq.cm = \(\frac{1728}{7}\) sq.cm = 389 \(\frac{5}{7}\) sq.cm.

Hence the whole surface area of the hemisphere =389 \(\frac{5}{7}\)sq.cm.

Class 10 Maths Solutions Wbbse

Example 11. A large new sphere is made by melting two metal spheres of radii rj unit and r2 unit respectively. Find the radius of the large sphere.

Solution:

Given:

A large new sphere is made by melting two metal spheres of radii rj unit and r2 unit respectively.

The total volume of the Mo spheres = \(\frac{4}{3}\)π \(\left(r_1^3+r_2^3\right)\) cubic. unit.

Now, let the radius of the large sphere be r unit.

∴ the volume of the large sphere = \(\frac{4}{3}\)πr3cu.umt.

As per condition,\(\frac{4}{3}\)πr3 = \(\frac{4}{3}\)πr3 \(\left(r_1^3+r_2^3\right)\)

⇒ r = \(\sqrt[3]{r_1^3+r_2^3}\)

Hence the required radius of the large sphere = \(\sqrt[3]{r_1^3+r_2^3}\) unit.

 

Mensuration Chapter 3 Sphere Long Answer Type Questions

Example 1. If the cost of making a leather ball is ₹431.20 at the rate of ₹17.50 per square cm. Calculate the length of diameter of the ball.

Solution:

Given:

If the cost of making a leather ball is ₹431.20 at the rate of ₹17.50 per square cm.

Let the radius of the ball = r cm.

∴ the whole surface area of the ball = 4πr2 sq-cm.

₹17.50 is required to make a ball of leather of 1 sq.cm

∴ ₹431.20 is required to make a ball of leather of \(\frac{431 \cdot 20}{17 \cdot 50}\) sq.cm

∴ 4πr2 = \(\frac{431 \cdot 20}{17 \cdot 50}\)

or, 4 x \(\frac{22}{7}\) x r2 = \(\frac{4312}{175}\)

or, \(r^2=\frac{4312 \times 7}{175 \times 4 \times 22}=\frac{49}{25}\)

or, \(r=\sqrt{\frac{49}{25}}=\frac{7}{5}\) =1.4

∴ the diameter of the ball = 2 x 1.4 cm = 2.8 cm.

Hence the diameter of the leather ball = 2.8 cm.

Example 2. If the length of diameter of a solid sphere is 28 cm and it is completely immersed into the water then calculate the volume of water displaced by the sphere.

Solution:

Given:

If the length of diameter of a solid sphere is 28 cm and it is completely immersed into the water

The diameter of the sphere = 28 cm

∴ radius of the sphere = \(\frac{28}{2}\) cm = 14 cm

∴ the volume of the sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x(14)3cc = 11498 \(\frac{2}{3}\) cc

Hence the required volume of water = 11498 \(\frac{2}{3}\) cc

Example 3. The length of radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it, then find the ratio of surface areas of the balloon in two cases.

Solution:

Given:

The length of radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it

The previous radius of the balloon = 7 cm.

∴ the previous surface area of the balloon = 4 x \(\frac{22}{7}\) x 72 sq.cm

∴ the ratio of the whole surface areas of the balloon in two cases

= 4 x \(\frac{22}{7}\) x 72: 4 x \(\frac{22}{7}\) x 212 = \(\frac{7 \times 7}{21 \times 21}\) = \(\frac{1}{9}\)= l: 9.

Hence the required ratio = 1:9.

Example 4. 127 \(\frac{2}{7}\)sq.cm of sheet is required to make a hemispherical bowl. Calculate the length of diameter of the forepart of the bowl.

Solution:

Given:

127 \(\frac{2}{7}\)sq.cm of sheet is required to make a hemispherical bowl.

Let the radius of the forepart of the bowl be r cm.

∴ the quantity of sheet required to make the bowl = 2πr2 sq.cm.

As per question, 2πr2=127 \(\frac{2}{7}\)

⇒ 2 x \(\frac{22}{7}\) x r2 = \(\frac{891}{7}\)

⇒ r2 = \(\frac{891 \times 7}{7 \times 2 \times 22}\)

⇒ r2 = \(\frac{81}{4}\)

⇒ r = \(\frac{9}{2}\)

So, the diameter of the bowl = 2 x \(\frac{9}{2}\) cm = 9cm.

Hence the required diameter of the forepart of the bowl = 9 cm.

Class 10 Maths Solutions Wbbse

Example 5. The length of diameter of a solid sphere of lead is 14 cm. If the sphere is melted, then calculate how many spheres with length of 3.5 cm radius can be made?

Solution:

Given:

The length of diameter of a solid sphere of lead is 14 cm. If the sphere is melted

The radius of the large solid sphere = \(\frac{14}{2}\) cm= 7cm

∴ volume of it = \(\frac{4}{3}\) x 73 cc.

The radius of each smaller sphere = 3.5 cm

∴ volume of each of it = \(\frac{4}{3}\)π x (3.5)3 cc

Let the number of smaller spheres which can be made is x.

∴ the volume of x smaller sphere = x x \(\frac{4}{3}\)π x (3.5)3 cc

As per condition, x x \(\frac{4}{3}\)π x (3.5)3 = \(\frac{4}{3}\)π x 73

⇒ x = \(\frac{7 \times 7 \times 7}{3 \cdot 5 \times 3 \cdot 5 \times 3 \cdot 5}\)

∴ x = 8

Hence the required number of smaller spheres = 8.

Example 6. Three spheres made of copper having the lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made. Calculate the length of radius of the large sphere.

Solution:

Given:

Three spheres made of copper having the lengths of 3 cm, 4 cm and 5 cm radii are melted and a large sphere is made.

The radii of three smaller spheres are 3 cm, 4 cm and 5 cm.

∴ the total volume of the three smaller spheres

= \(\frac{4}{3}\)π(33 + 43 +53)cc = \(\frac{4}{3}\)π(27 + 64 + 125)cc = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 216 cc.

Let the radius of the large sphere be R cm.

∴ volume ot the large sphere = \(\frac{4}{3}\) x \(\frac{22}{7}\) x R3 cc.

As per question, \(\frac{4}{3}\) x \(\frac{22}{7}\) x R3 = \(\frac{4}{3}\) x \(\frac{22}{7}\) x 216

⇒ R3 = 63

⇒  R = 6.

Hence the radius of the large sphere = 6 cm.

Example 7. The length of diameter of base of a hemispherical tomb is 42 dcm. Calculate the cost of colouring the upper surface of the tomb at the rate of 35 per square metre.

Class 10 Maths Solutions Wbbse

Solution:

Given:

The length of diameter of base of a hemispherical tomb is 42 dcm.

The diameter of base of the hemispherical tomb = 42 dcm

∴ its radius = \(\frac{42}{2}\) dcm = 21 dcm

∴ the surface area of the upper surface of the tomb

= 2 x \(\frac{22}{7}\) x 212 sq.dcm = 2772 sq.dcm = 27-72 sq. metres

So the cost of colouring the tomb = ₹27.72 x 35 = ₹970.2

Hence the required cost = ₹970.2.

Class 10 Maths Solutions Wbbse

Example 8. Two hollow spheres with the lengths of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal, Calculate the volumes of sheets of metal required to make the two spheres.

Solution:

Given:

Two hollow spheres with the lengths of diameter 21 cm and 17.5 cm respectively are made from the sheets of the same metal

The lengths of diameters of the two hollow spheres are 21 cm and 17.5 cm.

∴ their radii are \(\frac{21}{2}\) cm and \(\frac{17.5}{2}\) cm

So, the ratio of the curved surface areas of the two hollow spheres

= 4π x \(\left(\frac{21}{2}\right)^2\) : \(4 \pi \times\left(\frac{17 \cdot 5}{2}\right)^2\)

= \(\frac{21^2}{4}: \frac{(17 \cdot 5)^2}{4}\)

= 21 x 21: 17.5 x 17.5 = \(\frac{21 \times 21 \times 100}{175 \times 175}=\frac{36}{25}\) = 36: 25

Hence the required ratio = 36: 25

Example 9. The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere is half of that previous one. Calculate the ratio of the volumes of the portion cut off and remaining portion of the sphere.

Solution:

Given:

The curved surface of a solid metalic sphere is cut in such a way that the curved surface area of the new sphere is half of that previous one.

Let the radius of the metalic sphere be R unit and that of the new sphere produced be r unit.

As per question, 4πr2 = \(\frac{1}{2}\) x 4πR2

⇒ R2 = 2r2

⇒ R = √2r.

∴ the volume of the new sphere = \(\frac{4}{3}\)πr3 cubic.units.

Also, the volume of the portion cut off = \(\left(\frac{4}{3} \pi R^3-\frac{4}{3} \pi r^2\right)\) cubic.units.

= \(\frac{4}{3}\)πr3(R3-r3) cubic.units.

= \(\frac{4}{3}\) x \(\frac{22}{7}\)\(\left\{(\sqrt{2} r)^3-r^3\right\}\) cubic.units

= \(\frac{4}{3}\) x \(\frac{22}{7}\)(2√2r3 -r3) cubic.units

= \(\frac{4}{3}\) x \(\frac{22}{7}\)(2√2-1)r3 cubic, units.

∴ ratio of the volumes of the cut off the large sphere and the volume of the remaining part.

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x (2√2-1)r3 : \(\frac{4}{3}\)πr3 = (2√2-1):1

Hence the required ratio =(2√2-1) : 1.

Example 10. On the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes are made each of which has the length of radius 0.7 cm. Calculate the area of metal sheet surrounding its curved surface.

Solution:

Given:

On the curved surface of the axis of a globe with the length of 14 cm radius, two circular holes are made each of which has the length of radius 0.7 cm.

The whole surface area of the globe = 4π x (14)2 sq-cm.

The areas of the holes = 2π x (0.7)2 sq-cm.

So the required surface area of the two circular holes

= {4π x (14)2 – 2π x (0.7)2} sq-cm = 2π [2 x (14)2 – (0.7)2] sq-cm

= 2 x \(\frac{22}{7}\) x (392 – 0.49) sq.cm = 2 x \(\frac{22}{7}\) x 391.51 sq.cm = 2460.92 sq.cm.

Hence the required surface area of the circular metalic sheet = 2460.92 sq.cm.

Example 11. Calculate how many marbles with lengths of 1 cm radius may be formed by melting a solid sphere of iron having 8 cm length of radius.

Solution:

The radius of solid iron-sphere = 8 cm

∴ the volume of the iron-sphere = \(\frac{4}{3}\) x π x 83cc

The radius of each marble = 1 cm

∴ volume of each marble = \(\frac{4}{3}\)π x 13cc.

Let x marbles can be made.

∴ as per condition, x x \(\frac{4}{3}\)π x 13 = \(\frac{4}{3}\)π x 83

⇒ x = \(\frac{8^3}{1^3}\) = 512

Hence 512 marbles can be made.

Example 12. The external and internal radii of a hollow copper-sphere are 20 cm and 16 cm respectively. By melting this sphere, how many solid bullets of diameter 40 cm each can be made?

Solution:

Given:

The external and internal radii of a hollow copper-sphere are 20 cm and 16 cm respectively.

The external radius of the hollow sphere = \(\frac{20}{2}\) cm = 10 cm.

and the internal radius of the hollow-sphere = \(\frac{16}{2}\) cm = 8 cm

∴ the volume of materials of the hollow sphere

= \(\left[\frac{4}{3} \pi \times 10^3-\frac{4}{3} \pi \times 8^3\right]\) cc. = \(\frac{4}{3}\)π(1000- 512)cc = \(\frac{4}{3}\)π x 488cc

The radius of solid bullets = \(\frac{4}{2}\) = 2cm

∴ volume of solid bullets = \(\frac{4}{3}\)π x 23 cc.

Let the number of solid bullets that can be made be x.

∴ x x \(\frac{4}{3}\)3 x 23 = \(\frac{4}{3}\)π x 488

⇒ x = \(\frac{488}{8}\) = 61.

Hence 61 solid bullets can be made.

Class 10 Maths Solutions Wbbse

Example 13. If the radius of a sphere be increased by 3 cm, the volume of the sphere is increased by 264 cc. Find the radius of the sphere.

Solution:

Given:

If the radius of a sphere be increased by 3 cm, the volume of the sphere is increased by 264 cc.

Let the radius of the sphere be r cm.

∴ the volume of the sphere = \(\frac{4}{3}\)πr3 cc.

If the radius of the sphere be increased by 3 cm, then its volume will be \(\frac{4}{3}\)π(r + 3)3 cc.

As per question, \(\frac{4}{3}\)π(r + 3)3 – \(\frac{4}{3}\)πr3 = 264

or, \(\frac{4}{3}\)π[(r +3)3-r3] = 264 or, \(\frac{4}{3}\) x \(\frac{22}{7}\)(r3 +9r2 +27r + 27-r3) = 264

or, \(\frac{4}{3}\) x \(\frac{22}{7}\)(9r2+ 27r + 27) = 264

or, 9r2 +27r + 27 = \(\frac{264 \times 3 \times 7}{4 \times 22}\)

or, 9(r2 + 3r + 3) = 63 or, r2 + 3r + 3 = 7 or, r2 + 3r + 3 – 7 = 0 or, r2 + 3r- 4 = 0

or, r2 + 4r-r-4=0 or, r (r + 4) – 1 (r + 4) = 0 or, (r + 4)(r – 1) = 0

∴ either r + 4 = 0, or, r- 1 = 0 ⇒ r = – 4 or, r = 1

But the value of r cannot be negative, ∴ r = 1.

Hence the required radius of the sphere was 1 cm.

Example 14. If the radius of a solid sphere be decreased by 1 cm, then the curved surface area of it is decreased by 88 sq-cm. What was the radius of the square?

Solution:

Given:

If the radius of a solid sphere be decreased by 1 cm, then the curved surface area of it is decreased by 88 sq-cm

Let the radius of the solid sphere be r cm.

∴ the curved surface area of the sphere = 4πr2 sq-cm.

Now, if the radius of the sphere is decreased by 1 cm, then the curved surface of it will be 4 (r-1)2 sq-cm

As per question, 4πr2 – 4π(r- 1)2 = 88

or, 4π[r2 – (r – 1)2] = 88

or, 4 x \(\frac{22}{7}\)[(r+r-1)(r-r + 1)] = 88 or, 4 x \(\frac{22}{7}\)[(2r- 1)] = 88

or, 2r-1 = \(\frac{88 \times 7}{4 \times 22}\) or, 2r- 1 = 7 or, 2r = 8 or, r = 4.

Hence the radius of the solid sphere was 4 cm.

Example 15. The external diameter of a hollow gold-sphere is 12 cm and it is made of a plate of thickness 1 cm. If the mass of 1 cc gold be 19.5 gm and the price of 1 gm gold be X 1280, then what is the price of the whole solid sphere?

Solution:

Given:

The external diameter of a hollow gold-sphere is 12 cm and it is made of a plate of thickness 1 cm. If the mass of 1 cc gold be 19.5 gm and the price of 1 gm gold be X 1280

The external diameter of the gold-sphere is 12 cm and the thickness of it is 1 cm.

∴ the external radius of the gold-radius is \(\frac{12}{2}\)cm = 6 cm and its internal radius = (6- 1) cm = 5 cm.

So the volume of the sphere = \(\left(\frac{4}{3} \pi \times 6^3-\frac{4}{3} \pi \times 5^3\right)\) cc = \(\frac{4}{3}\)π(216-125) cc

= \(\frac{4}{3}\) x \(\frac{22}{7}\) x 91 cc = \(\frac{88 \times 13}{3}\) cc

The mass of 1 cc gold = 19.5 gm.

∴ the mass of \(\frac{88 \times 13}{3}\) cc gold = \(\frac{88 \times 13}{3}\) x 19.5 gm

∴ the price of the gold-sphere = ₹\(\frac{88 \times 13}{3}\) x 19.5 x 1280 = ₹9518080

Hence the required cost of the hollow gold-sphere = ₹9518080.

Example 16. Two solid spheres of radii 3 cm and 4 cm are melted to make a hollow sphere of external radius of 6 cm, then what will be the thickness of the new hollow sphere?

Solution:

Given:

Two solid spheres of radii 3 cm and 4 cm are melted to make a hollow sphere of external radius of 6 cm

The radii of the two solid spheres are 3 cm and 4 cm.

So, the tqtal volume of the two spheres

= \(\left(\frac{4}{3} \pi \times 3^3+\frac{4}{3} \pi \times 4^3\right)\)cc = \(\frac{4}{3}\) (27 + 64)cc = \(\frac{4}{3}\)π x 91 cc.

The external radius of the hollow sphere = 6 cm

Let the internal radius of it = r cm.

∴ the volume of material of the hollow sphere = \(\frac{4}{3}\)(63 -r3)cc = \(\frac{4}{3}\)(216-r3)cc

= \(\frac{4}{3}\)π(216-r3) = \(\frac{4}{3}\)π x 91 or, 216 – r3 = 91 or, r3 = 125 or, r3 = 53 ⇒ r = 5.

∴ thickness of the hollow sphere = (6 – 5) cm = 1 cm

Hence the thickness of the hollow sphere = 1 cm.

Class 10 Maths Solutions Wbbse

Example 17. The ratio of the volumes of two spheres is 216: 125. If the sum of the radii of the two spheres be 22 cm; then find the radii of the spheres.

Solution:

Given:

The ratio of the volumes of two spheres is 216: 125. If the sum of the radii of the two spheres be 22 cm

Let the radii of the two spheres be r1cm and r2 cm respectively.

As per question, \(\frac{4}{3} \pi r_1^3: \frac{4}{3} \pi r_2^3\) = 216: 125

⇒ \(\left(\frac{r_1}{r_2}\right)^3=\frac{216}{125}=\left(\frac{6}{5}\right)^3\)

⇒ \(\frac{r_1}{r_2}=\frac{6}{5}\)

⇒ \(r_1=\frac{6 r_2}{5}\)……(1)

Again, by question, r1 + r2 = 22

⇒ \(\frac{6 r_2}{5}+r_2\) = 22 [by (1)]

⇒ 11r2= 22 x 5

⇒ r2 = \(\frac{22 \times 5}{11}\) = 10

∴ from (1) we get, r1 = \(\frac{6 \times 10}{5}\) = 12

Hence the radii of the two spheres were 10 cm and 12 cm respectively.

Example 18. A sphere of greatest volume is cut off from a metal hemisphere of radius 6 cm. If the cost of 1 cc metal be ₹42, then what will be the cost of the remaining hemisphere?

Solution:

Given:

A sphere of greatest volume is cut off from a metal hemisphere of radius 6 cm. If the cost of 1 cc metal be ₹42

The radius of the metal hemisphere = 6 cm.

So, the volume of the hemisphere = \(\frac{2}{3}\)π x 63 cc.

Again, the diameter of the greatest sphere that can be cut off from the hemisphere will be 6 cm.

So, its radius will be \(\frac{6}{2}\)cm = 3cm.

∴ the volume of the greatest sphere cut off from the hemisphere = \(\frac{4}{3}\)π x 33 cc.

∴ The volume of the remaining part of the hemisphere

= \(\left(\frac{2}{3} \pi \times 6^3-\frac{4}{3} \pi \times 3^3\right)\)cc = \(\frac{2}{3}\)π(216-54)cc = \(\frac{2}{3}\)π x 162 cc = 108π cc.

So, the cost of the remaining part of the metal hemisphere = ₹108 x \(\frac{22}{7}\) x 42 = ₹14256.

Hence the required cost = ₹14256.

Example 19. The external diameter of the forepart of a bowl made of a copper plate of thickness I cm is 12 cm. If the weight of 1 cc of copper be 12 gm, then what will be the weight of the bowl?

Solution:

Given:

The external diameter of the forepart of a bowl made of a copper plate of thickness I cm is 12 cm. If the weight of 1 cc of copper be 12 gm

The external radius of the semicircular bowl = \(\frac{12}{2}\) cm = 6 cm.

The bowl is of thickness 1 cm.

So, the internal radius of the bowl = (6 – 1) cm = 5 cm.

∴ the volume of the material of the bowl

= \(\left(\frac{2}{3} \pi \times 6^3-\frac{2}{3} \pi \times 5^3\right)\)cc = \(\frac{2}{3}\)π(216-125) cc

= \(\frac{2}{3}\) x \(\frac{22}{7}\) x 91 cc

∴ the weight of the bowl = \(\frac{2}{3}\) x \(\frac{22}{7}\) x 91 x 21 gm = 2288 gm = 2.288 kg

Hence the weight of the semicircle bowl = 2.288 kg