Class 10 NCERT Solutions- Chapter 9 Some Application of Trigonometry - Exercise 9.1 | Set 2

Last Updated : 23 Jul, 2025

Chapter 9 of the Class 10 NCERT Mathematics textbook focuses on the "Some Applications of Trigonometry" exploring how trigonometric concepts can be applied to solve real-world problems. Exercise 9.1 | Set 2 is designed to reinforce students' understanding of these applications by providing practical problems involving heights and distances. This exercise is crucial for developing problem-solving skills and applying theoretical knowledge to practical scenarios.

Application of Trigonometry

In this chapter, students learn to use trigonometric ratios to solve problems related to heights and distances. This involves applying concepts such as the sine, cosine, and tangent functions to real-life situations like calculating the height of the building or the distance between two points. Understanding these applications is important for grasping how trigonometry is used in various fields such as engineering, architecture, and physics. This exercise helps students practice these concepts and develop their analytical skills.

The content of this article is merged in the article: Chapter 9 Some Application of Trigonometry - Exercise 9.1

Question 11. A TV tower stands vertically on the 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.). Find the height of the tower and the width of the canal.

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Solution:

In fig: AB = tower = ?
CB = canal = ?
In rt. ∆ABC,
tan60° =
h/x = √3
h = √3 x -(1)
In rt. ∆ABD
= tan 30°
= 1/√3
h = (x + 20)/√3 -(2)
From (1) and (2)
√3/1 = (x + 20)/√3
3x = x + 20
3x - x = 20
2x = 20
X = 20/2
X = 10
Width of the canal is 10m
Putting value of x in equation 1
h = √3 x
= 1.732(10)
= 17.32
Height of the tower 17.32m.

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.

Solution:

In fig: ED = building = 7m
AC = cable tower = ?
In rt ∆EDC,
= tan45°
7/x = 1/1
DC = 7
Now, EB = DC = 7m
In rt. ∆ABE,
= tan60°
AB/7 = √3/1
Height of tower = AC = AB + BC
7√3 + 7
= 7(√3 + 1)
= 7(1.732 + 1)
= 7(2.732)
Height of cable tower = 19.125m

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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.

Solution:

In fig:
AB = lighthouse = 75m
D and C are two ships
DC = ?
In rt. ∆ABD,
= tan30°
75/BD = 1/√3
BD = 75√3
In rt. ∆ABC
= tan45°
75/BC = 1/1
BC = 75
DC = BD - BC
= 75√3 - 75
75(√3 - 1)
75(1.372 - 1)
34.900
Hence, distance between two sheep is 34.900

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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 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.). Find the distance traveled by the balloon during the interval.

Solution:

In fig: AB = AC - BC
= 88.2 - 1.2
= 81m
In rt. ∆ABE
= 87/EB = tan30°
87/EB = 1/√3
EB = 87√3
In rt. ∆FDE
= tan60°
√3 ED = 87
ED = 87/√3
DB = DB - ED
87√3 - 87/√3
87(√3 - 1/√3)
= 87(3 - 1/√3)
= 87(2/√3) = 174/√3 * √3/√3
= 174 * √3/3 = 58√3
58 * 1.732 = 100.456m
Distance traveled by balloon is 100.456m

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.

Solution:

In fig: AB is tower
In rt. ∆ABD
= tan30°
= 1/√3
DB = √3 AB -(1)
In rt. ∆ABC
= tan60°
BC = AB/√3 -(2)
DC = DB - BC
= √3 AB - AB/√3
AB(3 - 1/√3)
CD = 2AB/√3
S1 = S2
\frac{D1}{T1} = \frac{D2}{T2}
2/√3AB/6 = AB/√3/t
2t = 6
t = 6/2
t = 3sec

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Question 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:

In fig: AB is tower
To prove: AB = 6m
Given: BC = 4m DB = 9m
In ∆ABC
= tanθ
AB/4 = tanθ -(1)
In ∆ABD
= tan (90°-θ)
AB/9 = 1/ tanθ
9/AB = tanθ -(2)
From (1) and (2)
AB/4 = 9/AB
AB² = 36
AB = √36
AB = √(6 * 6)
AB = 6m
Height of the tower is 6m.

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