Trigonometry helps students apply mathematical concepts to real-life situations involving heights, distances, and angles. NCERT Solutions for Class 10 Maths Chapter 9, Some Applications of Trigonometry, focuses on practical problems based on angles of elevation and depression, enabling students to solve height-and-distance questions with confidence. This chapter covers real-world applications such as towers, trees, buildings, kites, lighthouses, and roads, where trigonometric ratios are used to calculate unknown heights and distances.
Chapter 9 is highly important for the CBSE Class 10 board exams and also builds a strong connection with concepts introduced in Chapter 8 and used further in higher mathematics.
NCERT Solutions For Class 10 Maths Chapter 9 Some Applications Of Trigonometry
Q.
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 the following figure).
Q.
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.
Q.
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?
Q.
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.
Q.
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.
Q.
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.
Q.
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.
Q.
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.
Q.
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.
Q.
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 the following figure). Find the height of the tower and the width of the canal.
Q.
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.
Q.
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.
Q.
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 the following figure). Find the distance travelled by the balloon during the interval.
Q.
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.
Q.
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.
Class 10 Maths Chapter 9 Questions & Answers – Some Applications of Trigonometry
Questions & Answers
Q1. (Medium)
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:
Rope = 20 m (hypotenuse), angle with ground = 30°.
Height of pole = opposite side = 20 sin30°
= 20 × (1/2) = 10 m
Answer: 10 m
Q2. (Medium)
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 the tree break at height = x m from ground. Then distance on ground = 8 m and angle with ground = 30°.
tan30° = x/8 ⇒ x = 8/√3.
Now broken part (hypotenuse) = (H − x). Also, cos30° = 8/(H − x)
⇒ (H − x) = 8 ÷ (√3/2) = 16/√3.
So total height H = x + (H − x) = 8/√3 + 16/√3 = 24/√3 = 8√3 m
Answer: 8√3 m
Q3. (Medium)
A contractor plans to install two slides for children. For below 5 years: top height 1.5 m and angle 30°.
For elder children: height 3 m and angle 60°. Find length of slide in each case.
Solution:
(i) For height 1.5 m, angle 30°:
sin30° = 1.5/L ⇒ L = 1.5 ÷ (1/2) = 3 m
(ii) For height 3 m, angle 60°:
sin60° = 3/L ⇒ L = 3 ÷ (√3/2) = 6/√3 = 2√3 m
Answer: (i) 3 m, (ii) 2√3 m
Q4. (Medium)
The angle of elevation of the top of a tower from a point on the ground, 30 m away from the foot of the tower, is 30°.
Find the height of the tower.
Solution:
tan30° = h/30 ⇒ h = 30 × (1/√3) = 30/√3 = 10√3 m
Answer: 10√3 m
Q5. (Medium)
A kite is flying at a height of 60 m above the ground. The inclination of the string with the ground is 60°.
Find the length of the string (no slack).
Solution:
sin60° = 60/L ⇒ L = 60 ÷ (√3/2) = 120/√3 = 40√3 m
Answer: 40√3 m
Q6. (Medium)
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 increases from 30° to 60° as he walks towards the building.
Find the distance he walked.
Solution:
Height difference = 30 − 1.5 = 28.5 m.
Let initial distance = x, final distance = y.
tan30° = 28.5/x ⇒ x = 28.5√3.
tan60° = 28.5/y ⇒ y = 28.5/√3.
Distance walked = x − y = 28.5√3 − 28.5/√3 = 57/√3 = 19√3 m
Answer: 19√3 m
Q7. (Medium)
From a point on the ground, angles of elevation of the bottom and 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 distance from point to building = d.
Angle to bottom of tower (top of building) = 45°:
tan45° = 20/d ⇒ d = 20 m.
Angle to top of tower = 60°:
tan60° = (20 + h)/d = (20 + h)/20 = √3
⇒ 20 + h = 20√3 ⇒ h = 20(√3 − 1) m
Answer: 20(√3 − 1) m
Q8. (Medium)
A statue, 1.6 m tall, stands on the top of a pedestal.
From a point on the ground, angle of elevation of the top of the statue is 60° and to the top of the pedestal is 45°.
Find the height of the pedestal.
Solution:
Let pedestal height = p and distance from point = d.
tan45° = p/d ⇒ d = p.
tan60° = (p + 1.6)/d = (p + 1.6)/p = √3
⇒ p + 1.6 = √3 p ⇒ 1.6 = p(√3 − 1)
⇒ p = 1.6/(√3 − 1) = 0.8(√3 + 1) m
Answer: 0.8(√3 + 1) m
Q9. (Medium)
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 distance between tower and building = d, building height = h.
From tower foot: tan30° = h/d ⇒ h = d/√3.
From building foot: tan60° = 50/d ⇒ d = 50/√3.
So h = (50/√3)/√3 = 50/3 m
Answer: 50/3 m
Q10. (Medium)
A TV tower stands vertically on a bank of a canal.
From a point on the other bank directly opposite the tower, angle of elevation of the top 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 is 30°.
Find the height of the tower and the width of the canal.
Solution:
Let width of canal = w m, height of tower = h m.
From opposite point: tan60° = h/w ⇒ h = w√3.
Second point is 20 m farther on same line, so distance to tower foot = (w + 20).
tan30° = h/(w + 20) ⇒ h = (w + 20)/√3.
Equate: w√3 = (w + 20)/√3 ⇒ 3w = w + 20 ⇒ 2w = 20 ⇒ w = 10 m.
Then h = 10√3 m.
Answer: Height = 10√3 m, Width of canal = 10 m
Q11. (Medium)
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:
Let horizontal distance between building and tower = d, tower height = H.
Angle of depression to foot = 45° ⇒ tan45° = 7/d ⇒ d = 7 m.
Angle of elevation to top = 60° ⇒ tan60° = (H − 7)/d = (H − 7)/7 = √3
⇒ H − 7 = 7√3 ⇒ H = 7(√3 + 1) m
Answer: 7(√3 + 1) m
Q12. (Medium)
As observed from the top of a 75 m high lighthouse from 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:
Height of lighthouse = 75 m.
Distance of ship with 30° depression: d1 = 75/tan30° = 75√3.
Distance of ship with 45° depression: d2 = 75/tan45° = 75.
Distance between ships = d1 − d2 = 75√3 − 75 = 75(√3 − 1) m
Answer: 75(√3 − 1) m
Q13. (Medium)
A 1.2 m tall girl spots a balloon moving horizontally at a height of 88.2 m from the ground.
Angle of elevation is 60° and after some time reduces to 30°. Find the distance travelled by the balloon.
Solution:
Height from girl’s eyes = 88.2 − 1.2 = 87 m.
Let initial horizontal distance = x, later = y.
tan60° = 87/x ⇒ x = 87/√3 = 29√3.
tan30° = 87/y ⇒ y = 87√3.
Distance travelled = y − x = 87√3 − 29√3 = 58√3 m
Answer: 58√3 m
Q14. (Medium)
A straight highway leads to the foot of a tower. A man at the top observes a car at an angle of depression of 30°.
Six seconds later, the angle becomes 60°. Find the time taken by the car to reach the foot of the tower from this point.
Solution:
Let tower height = h. Let car distance initially = x and after 6 sec = y.
tan30° = h/x ⇒ x = h√3.
tan60° = h/y ⇒ y = h/√3.
Distance covered in 6 sec = x − y = h√3 − h/√3 = 2h/√3.
Speed = (2h/√3)/6 = h/(3√3).
Now from second position, remaining distance = y = h/√3.
Time = (h/√3) ÷ (h/(3√3)) = 3 sec
Answer: 3 seconds
Q15. (Medium)
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 and in the same straight line with it are complementary.
Prove that the height of the tower is 6 m.
Solution:
Let tower height = h. Let angles be α and β, and α + β = 90°.
From 4 m point: tanα = h/4.
From 9 m point: tanβ = h/9.
Since α and β are complementary: tanα · tanβ = 1.
So (h/4)(h/9) = 1 ⇒ h²/36 = 1 ⇒ h² = 36 ⇒ h = 6 m
Answer: Height of tower = 6 m