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.

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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°.

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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

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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).

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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.

 

 

 

 

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