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

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

WBBSE Class 10 Heights and Distances Overview

What is line of sight?

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.

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?

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.

What is angle of depression?

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.

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

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

Understanding Angles of Elevation and Depression

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

Solution:

Given

If the angle of elevation of the top of a tower from a distance of 10 metres from its foot is 60°,

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°

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.

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°

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.

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°

 

Solution:

Given

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

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.

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°

Solution:

Given

If the length of the shadow on the ground of a palm tree is √3 times of its height

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

Examples of Heights and Distances Problems

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

Solution: False

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

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:

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.

Common Formulas for Heights and Distances

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:

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:

Given

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 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:

In the right-angled triangle PQR, ∠Q = 90°, S is such a point on PQ that PQ: QR: QS = √3 : 1: 1

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°.

Applications of Trigonometry in Real-Life Heights and Distances

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:

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:

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:

Given

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.

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:

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:

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.)

WBBSE Class 10 Revision Notes on Heights and Distances

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:

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:

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:

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:

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:

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:

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:

Given

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°.

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.).

Word Problems on Heights and Distances with Solutions

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:

Given

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°.

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:

Given

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°.

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:

Given

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.

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:

Given

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

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:

Given

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.

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.

Visual Representation of Heights and Distances in Trigonometry

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:

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:

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:

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.

2. The two miles are at the points A and B respectively on the same, side of the observer.

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.