Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Exercise 9.1

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° (see Fig. 9.11).

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Solution

It can be observed that ABC is a right angled triangle with AC as the hypotenuse. Therefore,

sin C = AB/AC

sin 30° = AB/20

1/2 = AB/20

AB = 20/2 = 10 m

Hence, the height of the vertical pole is 10 m.

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

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let the remaining vertical part of tree be AB and the fallen part be AC.

It can be noted that ABC forms a right angled triangle with AC as its hypotenuse.

Thus,

cos C = BC/AC

cos 30° = 8/AC

√3/2 = 8/AC

AC = 16/√3 m

tan C = AB/BC

1/√3 = AB/8

AB = 8/√3 m

Height of the tree = AB + AC = 8/√3 m + 16/√3 m = 24/√3 m

Rationalise,

24/√3 × √3/√3 = 8√3 m

Hence, the height of the tree is 8√3 m.

3. A contractor plans to install two slides for the children to play in a park. For the 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 elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let the slide for children younger than 5 years be represented by ABC and the slide for elder children be represented by PQR.

AB = 1.5 m and PQ = 3 m.

In ABC, we have

sin C = AB/AC

sin 30° = 1.5/AC

1/2 = 1.5/AC

AC = 3 m

In PQR, we have

sin R = PQ/PR

sin 60° = 3/PR

√3/2 = 3/PR

PR = 6/√3 m

Rationalise,

6/√3 × √3/√3 = 2√3 m

Hence, the length of slide for younger children = 3 m,

and length of slide for elder children = 2√3 m

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the tower and C be the point 30 m away from the foot B.

tan C = AB/BC

tan 30° = AB/30

1/√3 = AB/30

AB = 30/√3 m

Rationalize,

30/√3 × √3/√3 = 10√3 m

Hence, the height of the tower is 10√3 m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let PQ be the height of the kite from ground and PR be the string attached to the kite.

sin R = PQ/PR

sin 60° = 60/PR

√3/2 = 60/PR

PR = 120/√3 m

Rationalise,

120/√3 × √3/√3 = 40√3 m

Hence, the length of the string assuming it has no slack is 40√3 m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let the above figure represent the given situation.

AE = Height of Building = 30 m

CF = Height of the boy = 1.5 m

BC = Distance between boy and the building

DC = Displacement of the boy from his original position C

In ABC, we have

tan C = AB/BC

tan 30° = (AE - CF)/BC

1/√3 = (30 - 1.5)/BC

BC = 28.5√3 m

In ABD, we have

tan D = AB/BD

tan 60° = 28.5/BD

√3 = 28.5/BD

BD = 28.5/√3 m

Rationalize,

28.5/√3 × √3/√3 = 9.5√3 m

CD = BC - BD = 28.5√3 m - 9.5√3 m

= 19√3 m

Hence, the boy walked 19√3 m towards the building.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the transmission tower on top of building BC. The required point on the ground is E.

In BCE, we have

tan CEB = BC/CE

tan 45° = 20/CE

1 = 20/CE

CE = 20 m

In ACE, we have

tan CEA = AC/CE

tan 60° = AC/20

√3 = AC/20

AC = 20√3 m

Height of transmission tower = AB = AC - BC = 20√3 - 20 = 20(√3 - 1) m

Hence, height of the transmission tower = 20(√3 - 1) m

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.

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Solution

Let AB be the statue on the pedestal BC and let E be the point on the ground.

In ACE, we have

tan CEA = AC/CE

tan 60° = (AB + BC)/CE

√3 = (1.6 + BC)/CE

(1.6 + BC)/√3 = CE

In BCE, we have

tan CEB = BC/CE

tan 45° = BC/CE

1 = BC/CE

CE = BC

Therefore,

(1.6 + BC)/√3 = BC

1.6 + BC = BC√3

1.6 = BC(√3 - 1)

BC = 1.6/(√3 - 1)

Rationlaize,

BC = 1.6/(√3 - 1) × (√3 + 1)/(√3 + 1)

BC = 1.6(√3 + 1)/2

BC = 0.8(√3 + 1) m

Hence, height of the pedestal is 0.8(√3 + 1) m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the building and DC be the tower.

In BCD, we have

tan B = DC/BC

tan 60° = 50/BC

√3 = 50/BC

BC = 50/√3

Rationalise,

BC = 50/√3 × √3/√3

= 50√3/3 m

In ABC, we have

tan C = AB/BC

1/√3 = AB/(50√3/3)

AB = 50/3 m

Hence, height of the building is 50/3 m.

10. Two poles of equal heights 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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB and DC be the poles on road BC and let the required point on the road be E.

In DCE, we have

tan E = CD/EC

tan 30° = CD/EC

1/√3 = CD/EC

CD = EC/√3

In ABE, we have

tan E = AB/BE

tan 60° = AB/BE

√3 = AB/BE

AB = BE√3

We know that AB = CD (Given). Therefore,

EC/√3 = BE√3

We also know that EC = 80 - BE. Therefore,

(80 - BE)/√3 = BE√3

80 - BE = 3BE

80 = 4BE

BE = 20 m

Therefore,

EC = 80 - 20 = 60 m

AB = CD = 20√3 m

Hence, the height of the poles is 20√3 m and the distance of the point from the poles is 20 m and 60 m respectively.

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 joing 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.

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Solution

In ABD, we have

tan D = AB/BD

tan 30° = AB/BD

1/√3 = AB/(20 + BC)

AB = (20+BC)/√3

In ABC, we have

tan C = AB/BC

tan 60° = AB/BC

√3 = AB/BC

AB = √3 BC

Therefore,

√3 BC = (20 + BC)/√3

3BC = 20 + BC

2 BC = 20

BC = 10 m

AB = 10√3 m

Hence, height of the tower is 10√3 m and width of the canal is 10 m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the 7 m high building and let the cable tower be EC. AD is an imaginary line drawn parallel to the ground BC.

AB = CD = 7 m

In ACD, we have

tan A = CD/AD

tan 45° = 7/AD

1 = 7/AD

AD = 7 m

In ADE, we have

tan A = DE/AD

tan 60° = DE/7

√3 = DE/7

DE = 7√3 m

EC = DE + CD = 7√3 + 7 = 7(√3 + 1) m

Hence, the height of the cable tower is 7(√3 + 1) m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the lighthouse and C and D be the positions of the ships in the sea respectively.

Draw an imaginary line AE which is parallel to the sea level BD.

In ABC, we have

tan C = AB/BC

tan 45° = AB/BC

1= 75/BC

BC = 75 m

In ABD, we have

tan D = AB/BD

tan 30° = AB/BD

1/√3 = 75/BD

BD = 75√3 m

CD = BD - BC = 75√3 - 75 = 75(√3 - 1) m

Hence, distance between the two ships in the sea = 75(√3 - 1) m

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.

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let A and B be the initial and final positions of the balloon. Let C be the head of the girl. AD and BE are imaginary lines drawn perpendicular to the ground from the positions of the balloon.

AD = BE = 88.2 - 1.2 = 87 m (Unchanged elevation in the balloon)

In ACD, we have

tan C = AD/CD

tan 60° = AD/CD

√3 = 87/CD

CD = 87/√3

Rationalize,

CD = 87/√3 × √3/√3 = 29√3 m

In BCE, we have

tan C = BE/CE

tan 30°= 87/CE

1/√3 = 87/CE

CE = 87√3 m

DE = CE - CD = 87√3 - 29√3 = 58√3 m

Hence, the distance travelled by the balloon is 58√3 m.

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.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the tower. Let D and C be the car's initial and final positions respectively.

Draw an imaginary line AE which is parallel to the ground.

In ABC, we have

tan C = AB/BC

tan 60° = AB/BC

√3 = AB/BC

BC = AB/√3

AB = √3 BC

In ABD, we have

tan D = AB/AD

tan 30° = AB/BD

1/√3 = AB/BD

AB = BD/√3

Rationalize,

AB = BD/√3 × √3/√3 = BD√3/3

Therefore,

BD√3/3 = √3 BC

BD = 3BC

3BC = BC + CD

2BC = CD

BC = CD/2

Time taken to travel distance CD = 6 s

Hence, time taken to travel distance BC = CD/2 = 6/2 = 3 s

16. The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution

Maths Solution Class 10 Chapter 9: Some Applications of Trigonometry

Let AB be the tower and C and D be the two required points on the ground.

Let angle ACB be x, then angle ABD = 90 - x

In ABC, we have

tan x = AB/BC

tan x = AB/4

AB = 4 tan x

In ABD, we have

tan (90°-x) = AB/BD

cot x = AB/9

AB = 9 cot x

Multiply both the obtained equations:

AB2 = 36 tan x cot x

AB2 = 36 tan x (1/tan x)

AB2 = 36

AB = 6 m as height cannot be negative.

Hence, the height of the tower has been proven to be 6 m.






Latest Courses