Chapter 9: Some Applications of Trigonometry (NCERT Solutions)

Exercise 9.1: Heights and Distances

Q1. 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°.

Let height of pole = h.
sin 30° = h/20.
1/2 = h/20.
h = 20/2 = 10 m.

Q2. 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.

Let broken part = x, standing part = h.
tan 30° = h/8 ⇒ 1/√3 = h/8 ⇒ h = 8/√3 = 8√3/3 m.
cos 30° = 8/x ⇒ √3/2 = 8/x ⇒ x = 16/√3 = 16√3/3 m.
Total height = h + x = 8√3/3 + 16√3/3 = 24√3/3 = 8√3 m.

Q3. 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?

Slide 1 (younger children):
sin 30° = 1.5/L₁.
1/2 = 1.5/L₁.
L₁ = 3 m.

Slide 2 (elder children):
sin 60° = 3/L₂.
√3/2 = 3/L₂.
L₂ = 6/√3 = 2√3 = 2√3 m ≈ 3.46 m.

Q4. 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.

tan 30° = h/30.
1/√3 = h/30.
h = 30/√3 = 10√3 m = 10√3 m ≈ 17.32 m.

Q5. 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.

sin 60° = 60/L.
√3/2 = 60/L.
L = 120/√3 = 40√3 = 40√3 m ≈ 69.28 m.

Q6. 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.

Effective height = 30 - 1.5 = 28.5 m.
Initially: tan 30° = 28.5/d₁ ⇒ d₁ = 28.5√3 m.
Finally: tan 60° = 28.5/d₂ ⇒ d₂ = 28.5/√3 = 9.5√3 m.
Distance walked = d₁ - d₂ = 28.5√3 - 9.5√3 = 19√3 = 19√3 m ≈ 32.91 m.

Q7. 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.

tan 45° = 20/d ⇒ d = 20 m.
tan 60° = (20+h)/d.
√3 = (20+h)/20.
20√3 = 20 + h.
h = 20√3 - 20 = 20(√3 - 1) = 20(√3 - 1) m ≈ 14.64 m.

Q8. 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.

tan 45° = h/d ⇒ d = h.
tan 60° = (h+1.6)/d.
√3 = (h+1.6)/h.
√3h = h + 1.6.
h(√3 - 1) = 1.6.
h = 1.6/(√3 - 1) = 1.6(√3 + 1)/2 = 0.8(√3 + 1) = 0.8(√3 + 1) m ≈ 2.19 m.

Q9. 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.

tan 60° = 50/d ⇒ √3 = 50/d ⇒ d = 50/√3 m.
tan 30° = h/d.
1/√3 = h/(50/√3).
h = 50/3 = 50/3 m ≈ 16.67 m.

Q10. 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.

Let distances be x and (80-x).
tan 60° = h/x ⇒ √3 = h/x ⇒ h = √3x.
tan 30° = h/(80-x) ⇒ 1/√3 = h/(80-x) ⇒ h = (80-x)/√3.
√3x = (80-x)/√3.
3x = 80 - x.
4x = 80 ⇒ x = 20 m.
h = √3(20) = 20√3 m ≈ 34.64 m.
Distances: 20 m and 60 m.

Q11. 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°. Find the height of the tower and the width of the canal.

tan 60° = h/x ⇒ h = √3x.
tan 30° = h/(x+20) ⇒ 1/√3 = h/(x+20) ⇒ h = (x+20)/√3.
√3x = (x+20)/√3.
3x = x + 20.
2x = 20 ⇒ x = 10 m.
h = √3(10) = 10√3 m ≈ 17.32 m.
Width of canal = 10 m.

Q12. 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.

tan 45° = 7/d ⇒ d = 7 m.
tan 60° = (h-7)/d.
√3 = (h-7)/7.
7√3 = h - 7.
h = 7√3 + 7 = 7(√3 + 1) = 7(√3 + 1) m ≈ 19.12 m.

Q13. 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.

tan 45° = 75/d₁ ⇒ d₁ = 75 m.
tan 30° = 75/d₂ ⇒ 1/√3 = 75/d₂ ⇒ d₂ = 75√3 m.
Distance = d₂ - d₁ = 75√3 - 75 = 75(√3 - 1) = 75(√3 - 1) m ≈ 54.9 m.

Q14. 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°. Find the distance travelled by the balloon during the interval.

Effective height = 88.2 - 1.2 = 87 m.
tan 60° = 87/d₁ ⇒ d₁ = 87/√3 = 29√3 m.
tan 30° = 87/d₂ ⇒ d₂ = 87√3 m.
Distance = d₂ - d₁ = 87√3 - 29√3 = 58√3 = 58√3 m ≈ 100.45 m.

Q15. 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.

tan 30° = h/d₁ ⇒ d₁ = h√3.
tan 60° = h/d₂ ⇒ d₂ = h/√3.
Distance covered in 6 sec = d₁ - d₂ = h√3 - h/√3 = h(3-1)/√3 = 2h/√3.
Speed = (2h/√3)/6 = h/(3√3).
Time to cover d₂ = (h/√3) / (h/(3√3)) = 3 sec.
Time = 3 seconds.

Q16. 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.

Let angles be θ and (90-θ).
tan θ = h/4.
tan(90-θ) = cot θ = h/9.
tan θ × cot θ = 1.
(h/4) × (h/9) = 1.
h²/36 = 1.
h² = 36.
h = 6 m. Proved.

Some Applications of Trigonometry - RD Sharma Important Questions

Q1. The angle of elevation of the top of a tower from a point on the ground is 30°. On walking 24 m towards the tower, the angle becomes 60°. Find the height of the tower.

tan 30° = h/(d+24) ⇒ h = (d+24)/√3.
tan 60° = h/d ⇒ h = √3d.
√3d = (d+24)/√3.
3d = d + 24 ⇒ 2d = 24 ⇒ d = 12.
h = 12√3 = 12√3 m ≈ 20.78 m.

Q2. A ladder 15 m long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

sin 30° = h/15 (angle with ground = 30°).
h = 15/2 = 7.5 m.

Q3. From a point on the ground, 40 m away from the foot of a tower, the angle of elevation of the top is 30°. The angle of elevation of the top of a water tank (on top of tower) is 45°. Find the height of the tank.

tan 30° = h_tower/40 ⇒ h_tower = 40/√3.
tan 45° = (h_tower+h_tank)/40 ⇒ h_tower+h_tank = 40.
h_tank = 40 - 40/√3 = 40(1 - 1/√3) = 40(√3-1)/√3 = 40(√3-1)/√3 m ≈ 16.92 m.

Q4. The shadow of a tower standing on a level ground is found to be 40 m longer when the sun's altitude is 30° than when it is 60°. Find the height of the tower.

tan 60° = h/x ⇒ x = h/√3.
tan 30° = h/(x+40) ⇒ x+40 = h√3.
h/√3 + 40 = h√3.
40 = h√3 - h/√3 = h(3-1)/√3 = 2h/√3.
h = 20√3 = 20√3 m ≈ 34.64 m.

Q5. From the top of a building 60 m high, the angles of depression of the top and bottom of a tower are 30° and 60° respectively. Find the height of the tower and its distance from the building.

tan 60° = 60/d ⇒ d = 60/√3 = 20√3 m.
tan 30° = (60-h)/d.
1/√3 = (60-h)/(20√3).
20 = 60 - h.
h = 40 m.
Height = 40 m, Distance = 20√3 m ≈ 34.64 m.

Q6. Two ships are sailing in the sea on the two sides of a lighthouse. The angles of elevation of the top of the lighthouse as observed from the two ships are 30° and 45° respectively. If the lighthouse is 100 m high, find the distance between the two ships.

tan 30° = 100/d₁ ⇒ d₁ = 100√3 m.
tan 45° = 100/d₂ ⇒ d₂ = 100 m.
Distance = d₁ + d₂ = 100√3 + 100 = 100(√3 + 1) = 100(√3 + 1) m ≈ 273.2 m.

Q7. The angle of elevation of a cloud from a point 60 m above a lake is 30° and the angle of depression of the reflection of the cloud in the lake is 60°. Find the height of the cloud.

Let cloud height above lake = H, observation point = 60 m.
tan 30° = (H-60)/d ⇒ H-60 = d/√3.
tan 60° = (H+60)/d ⇒ H+60 = d√3.
Divide: (H+60)/(H-60) = 3.
H + 60 = 3H - 180.
240 = 2H ⇒ H = 120 m.

Q8. A vertical tower stands on a horizontal plane and is surmounted by a flagstaff of height 7 m. At a point on the plane, the angle of elevation of the bottom of the flagstaff is 30° and that of the top of the flagstaff is 45°. Find the height of the tower.

tan 30° = h/d ⇒ d = h√3.
tan 45° = (h+7)/d ⇒ d = h+7.
h√3 = h + 7.
h(√3 - 1) = 7.
h = 7/(√3 - 1) = 7(√3+1)/2 = 7(√3+1)/2 m ≈ 9.56 m.

Q9. A man on the deck of a ship is 10 m above water level. He observes that the angle of elevation of the top of a cliff is 45° and the angle of depression of the base is 30°. Calculate the distance of the cliff from the ship and the height of the cliff.

tan 30° = 10/d ⇒ d = 10√3 m.
tan 45° = h/d ⇒ h = d = 10√3 m.
Height of cliff = 10 + h = 10 + 10√3 = 10(1+√3) m.
Distance = 10√3 m ≈ 17.32 m, Height = 10(1+√3) m ≈ 27.32 m.

Q10. The angles of depression of two consecutive kilometre stones on a straight road as seen from an aeroplane at an altitude of 1000 m are 45° and 30°. Find the distance of the aeroplane from the first stone and the speed of the plane if it takes 10 seconds to fly between the two stones.

tan 45° = 1000/d₁ ⇒ d₁ = 1000 m.
Distance from first stone = √(1000²+1000²) = 1000√2 m.
tan 30° = 1000/d₂ ⇒ d₂ = 1000√3 m.
Horizontal distance covered = d₂ - d₁ = 1000(√3-1) m.
Speed = 1000(√3-1)/10 = 100(√3-1) = 100(√3-1) m/s ≈ 73.2 m/s.

Q11-Q20. [Additional important problems covering various scenarios with angles of elevation/depression, including problems on: flag posts, balloons, aircraft, roads on hills, and combined height-distance problems]

These follow similar solution approaches using:
• tan θ = height/distance
• Complementary angles property
• Angle of depression = angle of elevation from base
• Breaking complex problems into multiple right triangles

Some Applications of Trigonometry - Formulas & PYQs

Key Concepts & Formulas

1. Line of Sight

The line from the observer's eye to the object being observed.

2. Angle of Elevation

The angle formed by the line of sight with the horizontal when the object is above the horizontal level.

tan(θ) = Height / Horizontal Distance

3. Angle of Depression

The angle formed by the line of sight with the horizontal when the object is below the horizontal level.

Key: Angle of depression from point A to B = Angle of elevation from point B to A

4. Common Problem Types

Type 1: Finding height when distance and angle are known
h = d × tan(θ)

Type 2: Finding distance when height and angle are known
d = h / tan(θ)

Type 3: Two angles from different positions
Form two equations and solve simultaneously

Previous Year Questions (CBSE/JKBOSE)

Q1. 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. (CBSE 2020)

tan 30° = h/30.
1/√3 = h/30.
h = 30/√3 = 10√3 m = 10√3 m.

Q2. From a point on the ground, the 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. (CBSE 2019)

Standard NCERT Q7. Height = 20(√3 - 1) m.

Q3. The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun's altitude is 30° than when it was 60°. Find the height of the tower. (CBSE 2018)

h = 20√3 m.

Q4. An aeroplane when flying at a height of 4000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are 60° and 45° respectively. Find the vertical distance between the aeroplanes at that instant. (CBSE 2017)

tan 45° = 4000/d ⇒ d = 4000 m.
tan 60° = h₂/d ⇒ h₂ = 4000√3 m.
Vertical distance = 4000√3 - 4000 = 4000(√3 - 1) = 4000(√3 - 1) m.

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