NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry
Last Updated :
23 Jul, 2025
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry - This article has been designed and reviewed by subject experts at GFG as a tool to assist students in resolving questions related to Applications of Trigonometry.
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- NCERT Maths Solutions Class 10 Chapter 9 Some Applications of Trigonometry Exercise 9.1 Set 1, Set 2 – 16 Questions (16 Long Answers)
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The study of how trigonometry is used practically in daily life is the subject of this chapter. The line of sight, angle of deviation, angle of elevation, and angle of depression are all covered in this topic. The chapter's exercise is based on the discussion of how trigonometry may be used to determine the heights and distances of various objects.
The solutions to all the exercises in NCERT Chapter 9 Some Applications of Trigonometry have been covered in the NCERT Solutions for Class 10 Maths.
NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry Exercise 9.1
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° (see Fig.).
Solution:
In rt ∆ABC,
AB = pole = ?
AC = rope = 20m
sinθ = \frac{P}{H}
sin30° = \frac{AB}{AC}
AB = 1/2 * 20
AB = 10m
Height of pole = 10m
Question 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:
In rt ∆ABC,
BC = 8m
\frac{P}{B} = \frac{AB}{BC}
= tan30°
\frac{AB}{BC}
= 1/√3
AB = 8/√3 -(1)
Now,
\frac{BC}{ AC}
= cos30°
8/AC = √3/2
√3AC = 16
AC = 16/√3 -(2)
From (1) and (2)
Height of tree = AB + AC
= 8/√3 * 16√3
= 8√3 m
8 * 1.73 = 13.84m
The height of the tree is 13.84
Question 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 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
Solution:
In rt ∆ABC,
AB = 1.5m
AC = side = ?
\frac{P}{H} = \frac{AB}{AC}
= sin30°
1.5/AC = 1/2
AC = 1/5 * 2
AC = 3m
In rt ∆PQR,
PQ = 3m
PR = side = ?
\frac{P}{H} = \frac{PQ}{PR}
= sin60°
3/PR = √3/2
√3 PR = 6
PR = 6/√3
6/√3 * √3/√3
= 2√3
= 2 * 1.73
= 3.46m
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.
Solution:
In rt ∆ABC,
AB = tower = ?
BC = 30m
\frac{P}{B} = \frac{AB}{BC}
= tan30°
AB/30 = 1/√3
AB = 30/√3
AB = 30/√3 * √3/√3
= (30√3)/3 = 10√3
= 10 * 1.73
= 17.3m
The height of tower 17.3m
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.
Solution:
In rt ∆ABC,
AB = 6Om
AC = string = ?
\frac{P}{H} = \frac{AB}{AC}
= sin60°
60/AC = √3/3
√3 AC = 60 * 2
AC = 120/120/(√3) * √3/√3
120/√3 * √3/√3
40 = √3
40 * 1.73 = 69.20m
Length of the string is 69.20m
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.
Solution:
In fig AB = AE - 1.5
= 30 - 1.5
= 28.5
In rt ∆ABD,
\frac{P}{B} = \frac{AB}{BD}
= tan30°
= 28.5/BD = 1/√3
BD = 28.5√3 -(1)
In rt ∆ABC,
\frac{P}{B} = \frac{AB}{BC}
= tan60°
28.5/BC*√3
√3 BC = 28.5
BC = 28.5/√3 -(2)
CD = BD − BC
= 28.5√3 - 28.5/√3
= 28.5(2/√3)
57/√3 * √3/√3 = (57√3)/3 = 19√3
19 * 1.73 = 32.87m
The boy walked 32.87m towards the building.
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.
Solution:
In fig:
AB = tower = ?
BC = building = 20m
In rt ∆BCD
\frac{P}{B} = \frac{BC}{CD}
= tan45°
20/CD = 1/1
CD = 20
In rt. ∆ACD,
\frac{P}{B} = \frac{AC}{CD}
= tan60°
AC/20 = √3/1
AC = 20√3 -(1)
AB = AC-BC
20√3 - 20
20(√3 - 1)
20(1.732 - 1)
20(0.732)
14.64m
The height of the tower is 14.6m
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.
Solution:
In fig: AB = statue = 1.6m
BC = pedestal = ?
In rt ∆ACD
\frac{P}{B} = \frac{AC}{CD}
= tan60°
1.6 + BC/CD = √3
√3 CD = 1.6 + BC
CD = 1.6+BC/√3 -(1)
In rt ∆BCD,
\frac{BC}{CD}
= tan45°
\frac{BC}{CD}
= 1/1
CD = BC
From (1)
1.6 + BC/√3 = BC/1
√3 BC = 1.6 + BC
1.732 BC - 1 BC = 1.6
0.732 * BC = 1.6
BC = 1.6/0.732
BC = 16/10 * 100/732 = 1600/732
BC = 2.18m
Height of pedestal is 2.18m
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.
Solution:
In fig:
AB = tower = 50m
DC = building = ?
In rt.∆ABC,
\frac{P}{B} = \frac{AB}{BC}
= tan60°
√3 BC = 50
BC = 50/√3
In rt. ∆DCB
\frac{P}{B} = \frac{DC}{BC}
= tan30°
\frac{P}{B} = \frac{DC}{\frac{50}{√3}}
= 1/√3
DC = 50/√3
DC = 50/√3 * 1/√3
DC = 50/3
DC = 16\tfrac{2}{3}
The height of the building is 16\tfrac{2}{3}
m
Question 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:
AB and CD on equal poles.
Let their height = h
Let DP = x
Then PB = BD - x
In rt. ∆CDP,
\frac{P}{B} = \frac{CD}{DP}
= tan60°
h/x = √3/1
h = √3 x -(1)
In rt. ∆ABP
\frac{P}{B} = \frac{AB}{BP}
= tan30°
h/(80 - x) = 1/√3
h = (80 - x)/√3 -(2)
From (1) and (2)
(√3 x)/1 = 80 - x/√3
3x = 80 - x
3x + x = 80
4x = 80
X = 80/4
X = 20
Putting values of X in equation 1
h = √3 x
h = √3(20)
h = 1.732(20)
h = 34.640
Height of each pole = 34.64m
The point is 20m away from first pole and 60m away from second pole.
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.). Find the height of the tower and the width of the canal.
Solution:
In fig: AB = tower = ?
CB = canal = ?
In rt. ∆ABC,
tan60° = \frac{AB}{BC}
h/x = √3
h = √3 x -(1)
In rt. ∆ABD
\frac{AB}{BD}
= tan 30°
\frac{h}{x + 20}
= 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,
\frac{ED}{DC}
= tan45°
7/x = 1/1
DC = 7
Now, EB = DC = 7m
In rt. ∆ABE,
\frac{AB}{BC}
= 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
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,
\frac{AB}{BD}
= tan30°
75/BD = 1/√3
BD = 75√3
In rt. ∆ABC
\frac{AB}{BC}
= 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
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
\frac{AB}{EB}
= 87/EB = tan30°
87/EB = 1/√3
EB = 87√3
In rt. ∆FDE
\frac{FD}{ED}
= 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
\frac{AB}{DB}
= tan30°
\frac{AB}{DB}
= 1/√3
DB = √3 AB -(1)
In rt. ∆ABC
\frac{AB}{BC}
= tan60°
BC = AB/√3 -(2)
DC = DB - BC
= √3 AB - AB/√3
AB(3 - 1/√3)
CD = 2AB/√3
Speed = \frac{Distance}{Time}
S1 = S2
\frac{D1}{T1} = \frac{D2}{T2}
\frac{DC}{6} = \frac{CB}{t}
2/√3AB/6 = AB/√3/t
2t = 6
t = 6/2
t = 3sec
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
\frac{AB}{BC}
= tanθ
AB/4 = tanθ -(1)
In ∆ABD
\frac{AB}{BD}
= 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.
Key Features of NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry:
- These NCERT solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry have been developed by the GfG team, with a focus on students’ benefit.
- These solutions are entirely accurate and can be used by students to prepare for their board exams.
- Each solution is presented in a step-by-step format with comprehensive explanations of the intermediate steps.
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