# Some Applications of Trigonometry – Exercise 9.1-Class 10

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 °. Solution:

It can be observed from the figure that AB is the pole.

In ∆ABC,

AB/AC = sin 30°

AB/20  = 1/2

AB = 20/2  = 10

Therefore, the height of the pole is 10 m.

1. 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: Let AC was the original tree. Due to storm, it was broken into two parts.

The broken part AB is making 30° with the ground.

In ∆A’BC,

BC/A’C = tan 30°

BC/8  = 1/√3

BC = (8/√3)m

A’C/A’B = cos 30°

8/A’B  = √3/2

A’B = (16/√3)m

Height of the tree = A’B + BC

= (8/√3 + 16/√3)m = 24/√3 m = 8√3m

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

1. 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, where as for the elder children she wants to have a steep side 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?

Solution:

It can be observed that AC and PR are the slides for younger and elder children respectively. In ∆ABC,

AB/AC = sin30°

1.5/AC  = 1/2

AC = 3 m In ∆PQR,

PQ/PR = sin60°

3/PR  = √3/2

PR = 6/3 = 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: Let AB be the tower and the angle of elevation from point C is 30°.

In ∆ABC,

AB/AC = tan30°

AB/30  = 1/√3

AB = 30/3 = 10√3m

Therefore, 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: Let K be the kite and the string is tied to point P on the ground.

In ∆KLP,

KL/KP = sin30°

60/KP  = √3/2

KP = 120/3 = 40√3m

Hence, the length of the string 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: Let the boy was standing at point S initially. He walked towards the building and reached at point T.

PR = PQ − RQ = (30 − 1.5) m = 28.5 m = 57/2 m

In ∆PAR,

PR/AR = tan30°

57/AR  = 1/√3

AR = (57/2 √3) m

In ∆PRB,

PR/BR = tan60°

57/2 BR  = √3

BR = 56/2√3 = m

ST = AB = AR – BR = (57/2 √3 – 19√3/2)m = 38/2 √3 m = 19 √3 m

Hence, he 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: Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In ∆BCD,

BC/CD = tan45°

20/CD  = 1

CD = 20 m

In ∆ACD,

AC/CD = tan60°

AB+BC/CD  = √3

AB = (20√3 – 20)m

= 20(√3 – 1)m

Therefore, the height of the transmission tower is 20(√3 − 1) m.

8. A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of 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.

Solution: Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In ∆BCD,

BC/CD = tan45°

BC/CD = 1

CD = BC

In ∆ACD,

AB+BC/CD = tan60°

AB+BC/CD  = √3

1.6 + BC = BC√3

BC(√3 – 1)= 1.6

BC = (1.6)( √3+1)/(√3-1)( √3+1)

= (1.6)( √3+1)/(√3)2– (1)2

= 1.6(√3+1)/2 = 0.8(√3+1)

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

1. 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: Let AB be the building and CD be the tower. In ∆CDB,

CD/BD = tan60°

50/BD  = √3

BD = 50/3 m

In ∆ABD,

AB/BD = tan30°

AB/CD  = 50/√3 X 1/√3 = 50/3 = 16 2/3

Therefore, the height of the building is 16 2/3 m.