NCERT Mathematics class X SOme Application of trigonometry

 SOME APPLICATIONS OF TRIGONOMETRY

What is the angle of elevation? 

 When a person sees an object that lies at a height above the ground, the angle between the horizontal level and the line joining the object from the observer is called the angle of elevation.

What is the angle of depression? 

 When an observer sees an object lying on the ground from a height, then the angle between the horizontal line (from the observer's eye level) and the line joining the object and the eye is said to be the angle of depression.

In the above figure, the first figure is the angle of elevation, and the second figure is the angle of depression. 

Simple rules for calculation

Unknown value/Known value = Trigonometric ratio

Solution of Question 1

A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is 30°.

Here, given that the Rope AC is 20 m long, AB is the pole. The height of the pole AB is unknown. The angle of elevation from C to A is 30 degrees

AB/AC = Sin30° 

AB = AC Sin 30° 

       = 20*1/2 = 10 m

The height of the pole is 10 m

Question 2

A tree breaks due to a storm, and the broken part bends so that the top of the tree touches the ground, making an angle of 30° with it. The distance between the foot of the tree and the point where the top touches the ground is 8 m. Find the height of the tree.

The tree is broken from point A and touches the ground at point C. We have to find the AB and AC so that the total height of the tree is = AC

AB/BC = tan 30° 

AB = BC tan 30° 

AB = 8 * 1/厂3  = 8/厂3 m 

AC/BC = Sec 30° 

AC = BC*2/厂3 m

AC = 16/厂3 m

Total height of the tree = AB + AC = 8/√3 m + 16/√3 m

                                      = 24/√3 m 

                                      = 8*√3*√3/√3 = 8√3 m

Question 3

         A contractor plans to install two slides for the children to play on in a park. For children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m and is inclined at an angle of 30° to the ground, whereas for older children, she wants to have a steep slide at a height of 3 m and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

The first triangle, ABC, shows the slide for the 5-year-old children, and the second figure, triangle PQR, shows the slide for the older children.

We have to find AC and PR, which are the lengths of the sides.

In triangle ABC

AC/AB = Cosec30°

AC = ABcosec30°

       = 1.5* 2 = 3 m

In the triangle PQR 

PR/PQ = cosec 60°

PR = PQ cosec 60°

      = 3*2/√3   

= √3*√3*2/√3 

= 2√3 m 

Question 4:

The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Given that the distance BC = 30m

Angle C = 30°

So, AB/BC = tan 30°.

AB/30 = 1/√3

AB = 30/√3 = 10*√3*√3/3 = 10√3 m

The height of the tower is 103 m 

Question 5

A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

In the figure, the height of the kite from the ground is 60 m, represented by RQ we want to find the length of the string RP

The angle of elevation of the kite from the ground is 60°

PR/PQ = Cosec60°

PR = PQ Cosec 60°

= 60* 2/√3 = 10*2*√3 *√3 *2/ √3 = 40√3 m. 

Question 6:

A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

PQ is a tower of height 30 m. AB is a boy of height 1.5 m.

AB = RQ = 1.5m

RP = 30-1.5 = 28.5 m

When the boy was at A, the angle of elevation was 30°.

When he reaches point C, the angle becomes 60°.

Consider the 🔼PRA

AR/PR = cot 30°

AR  = PR cot 30°

      = 28.5✓3 m

In the 🔼PRC

RC/PR = cot 60°

RC = PRcot60°

     = 28.5*1/√3

     = 28.5√3/3  

Distance moved by the boy 

AC = RA - RC

 = 28.5√3 m - 28.5√3/3

= 28.5√3 (1-1/3) = 28.5√3 *2/3 = 57√3/3 = 19√3 m

Question 7

From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60°, respectively. Find the height of the tower.

Given that the building height = 20 m = AB

Tower height = BC [to find out]

AD = distance of the point from the foot of the building A

In  🔼ABD

AD/AB = cot45° = 1

AD = AB =20m

Now, in 🔼ACD

AC/AD = tan 60°

AC = AD tan 60° = 20√3

Height of the tower = AC - AB 

                                = 20√3 - 20

                                =20(√3-1) m

Question 8:

A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60°, and from the same point, the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

AB = statue height = 1.6m

BC = pedestal

CD is the distance of the point D from the pedestal

In 🔺BCD

BC/CD = tan 45° = 1

BC = CD   [1]

 [height of pedestal = distance of point D from pedestal]

In 🔺ACD

AC/CD = tan 60° = √3 

AC = CD√3 

AB + BC = CD√3 

Since BC = CD from [1]

1.6 + BC = BC√3 

1.6 = BC√3 - BC

1.6 = BC(√3 - 1)

1.6/(√3 -1) = BC

We can rationalize to simplify it by multiplying the numerator and denominator by √3 + 1.

BC = 1.6(√3 +1)/2 = 0.8(√3 +1) m

Height of the pedestal = 0.8(√3 + 1) m

Question 9:

The angle of elevation of the top of a building from the foot of the tower is 30°, and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

CD = height of the tower = 50m

AB is the height of the building.

Angle CBD = 60°

Angle ADB = 30°

In the triangle CBD 

BD/CD = Cot 60√3

BD = CD 60°

      = 50*1/√3

Now in 🔼ABD

AB/BD = tan 30°

AB = BD tan 30°

AB  = 50 * 1/√3 * 1/√3 = 50/3 m

Height of the building = 50/3 m

Question 10

Two poles of equal height are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Suppose P is the point on the ground from where the angle of elevation of the poles on either side of the Road BD (80m) is 60° and 30°
Let BP = x m

and PD = (80-x ) m

In the triangle ABP

AB/BP = tan 60°

AB = BP tan 60°

       = x√3 m

In triangle CDP

CD/PD = tan 30°

CD = (80-x)*1/√3

Since the height AB = CD

therefore,

x√3 m = (80-x)*1/√3

3x = 80 - x

4x = 80

x = 80/4 = 20m

PD = 80-20 = 60m

Height of the poles = 20√3m

Question 11

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Given that the TV tower AB makes an angle of 30° with the point D and 60° with the point B

CD = 20m

Let BC = x m

In triangle ABC 

AB/BC = tan 60°

AB = BC√3  m             [1]

In the triangle ABD

AB/BD = tan 30°

AB = BD √3 = (20+BC)/√3

from [1]

BC√3 = (20+BC)/√3

3BC = 20 + BC

2BC = 20 

BC = 10 m

Therefore,

AB = BC√3  m   = 20√3 m  

Question 12

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60°, and the angle of depression of its foot is 45°. Determine the height of the tower. 

AB = building of height = 7 m

AB = ED = 7 m

EC =?

Angle of elevation from A to C = 60°

The angle of depression from A to D is 45°

In the triangle ABD

AB/BD = tan45° = 1

AB = BD = 7 m

In the triangle AEC

CE/AE = tan 60°

CE = AE tan 60° = AE√3 [AE = BD]

CE = 7√3m

Hence, the height of the Tower CD = EC + ED = EC + AB = (7 + 7√3) m.

Question 13

As observed from the top of a 75 m high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

AB is the lighthouse of 75 m height. Two ships are at positions P and Q. 

The angle of elevation of the Ship at position P and Q are 30° and 60°

In the ▲ABQ

BQ/AB = Cot 45°

BQ/AB = 1

BQ = AB = 75m

In 🔼ABP

BP/AB  = cot 30°

BP = AB*√3

BP = 75√3 

Rationalising

BP = 75√3 m

BQ + QP = 75√3 m

QP = 75√3 - 75

      = 75(√3-1) m 

Distance between the ships = 75(√3-1) m

Question 14

A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval. 

In this figure the balloon moved from place A to C; the height of the balloon from the ground is 88.2 m.

PQ = the height of a girl = 1.2 m

The height of the balloon from the eye level of the girl = AB = CD = 88.2-1.2 = 87m

Angle APB = 60°, Angle CPD = 30°

In the 🛆APB

PB/AB = cot 60°

PB = AB * 1/⇃3 = 87 * 1/⇃3

In  🛆CPD

PD/CD = cot 30°

PD = CD ⇃ 3 = 87 ⇃ 3 m

PB + BD = PD

87/⇃3 + BD = 87 ⇃ 3

BD = 87 ⇃ 3 - 87/⇃3

       = 87(⇃3-1/⇃3) = 87*2/⇃3

       = 87*2*⇃3/3 [rationalizing]

       =58⇃3 m

BD is the distance covered by the balloon. 

Question 15

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point. 

    

BD is a highway, AB is a tower. Man standing at the top of the tower obsers the angle of depression for the poin D and C are 30° and 60° in 6 sec  duration.  AX is horizontal level from eye.

therefore angle D and C will be alternate angles formed between the parallel lines AX and BD, cut by the transversal AD and AC

In 🔺ABC

BC/AB = cot 60°

BC= AB/⇃3  ----[1]

In 🔺ABD 

BD/AB = cot30°

BD = AB√3

 CD = BD -BC =AB√3 - AB/⇃3 = AB(3-1)/⇃3

CD = 2AB/⇃3

CD/BC = 2AB/⇃3/AB√3 = 2

CD = 2BC

Let the speed of the car is v

Then the time taken for the CD = distance/speed = 2BC /v

Time taken to reach the foot of the tower = distance/speed 

                                                                   = BC/v

Ratio of time  = time for BC/ time for CD

                      =BC/v/2BC /v = 1/2

time for BC/ time for CD = 1/2

Time for BC = Time for CD/2 = 6sec/2 = 3sec

CBSE BOARD QUESTIONS

A man on a cliff observes a boat at an angle of depression of 30º which is approaching the shore to the point immediately beneath the observer with a uniform speed. Six minutes later, the angle of depression of the boat is found to be 60º. Find the time taken by the boat form here to reach the shore.    (CBSE 2024)

Solution 

 This question is just like Question 15, only the words are changed and car is replaced by the boat. tower is replaced by the cliff. 

Question2. If a pole 6 m high casts a shadow 2√3 m long on the ground, then sun's elevation is         (CBSE 2023) 

(a) 60º (b) 45º (c) 30º (d) 90º

***Contains the solution of  NCERT MATH of CLASS 10***

*For the question related to board you can send your query.