NCERT Exercise Questions & Answers + Important Extra Questions
Exercise 9.1
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°.
Answer
sin 30° = height/rope = h/20
⇒ 1/2 = h/20
⇒ h = 10 m
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of 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.
Answer
Let broken part = l, standing part = h. tan 30° = h/8
⇒ h = 8/√3. cos 30° = 8/l
⇒ l = 8/(√3/2) = 16/√3.
Total height = h + l = 8/√3 + 16/√3 = 24/√3 = 24√3/3 = 8√3 m ≈ 13.86 m
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5, 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?
Answer
For younger children: sin 30° = 1.5/l₁
⇒ 1/2 = 1.5/l₁
⇒ l₁ = 3 m
For elder children: sin 60° = 3/l₂
⇒ √3/2 = 3/l₂
⇒ l₂ = 6/√3 = 2√3 m ≈ 3.46 m
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.
Answer
tan 30° = h/30
⇒ 1/√3 = h/30
⇒ h = 30/√3 = 10√3 m ≈ 17.32 m
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.
Answer
sin 60° = 60/l
⇒ √3/2 = 60/l
⇒ l = 120/√3 = 40√3 m ≈ 69.28 m
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.
Answer
Effective height = 30 – 1.5 = 28.5 m.
From initial position: tan 30° = 28.5/d₁
⇒ d₁ = 28.5√3
From final position: tan 60° = 28.5/d₂
⇒ d₂ = 28.5/√3 = 9.5√3
Distance walked = d₁ – d₂ = 28.5√3 – 9.5√3 = 19√3 m ≈ 32.91 m
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.
Answer
Let distance from building = d. tan 45° = 20/d
⇒ d = 20 m.
tan 60° = (20+h)/20
⇒ √3 = (20+h)/20
⇒ 20+h = 20√3
⇒ h = 20(√3–1) = 20(√3–1) m ≈ 14.64 m
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.
Answer
Let pedestal height = h, distance = d. tan 45° = h/d
⇒ d = h.
tan 60° = (h+1.6)/d = (h+1.6)/h
⇒ √3 = 1 + 1.6/h
⇒ 1.6/h = √3–1
⇒ h = 1.6/(√3–1) = 1.6(√3+1)/2 = 0.8(√3+1) m ≈ 2.19 m
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.
Answer
Let building height = h, distance between them = d.
tan 60° = 50/d
⇒ d = 50/√3. tan 30° = h/d
⇒ 1/√3 = h/(50/√3)
⇒ h = 50/3 = 50/3 m ≈ 16.67 m
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.
Answer
Let height = h, distances from point = x and (80–x).
tan 60° = h/x
⇒ h = x√3. tan 30° = h/(80–x)
⇒ h = (80–x)/√3.
x√3 = (80–x)/√3
⇒ 3x = 80–x
⇒ 4x = 80
⇒ x = 20.
h = 20√3 = 20√3 m ≈ 34.64 m. Distances: 20 m and 60 m.
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 is 30°. Find the height of the tower and the width of the canal.
Answer
Let height = h, width = d. tan 60° = h/d
⇒ h = d√3. tan 30° = h/(d+20)
⇒ h = (d+20)/√3.
d√3 = (d+20)/√3
⇒ 3d = d+20
⇒ 2d = 20
⇒ d = 10. h = 10√3.
Height = 10√3 m ≈ 17.32 m, Width = 10 m
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.
Answer
Let distance = d. tan 45° = 7/d
⇒ d = 7. Let extra height of tower above building = h.
tan 60° = h/7
⇒ h = 7√3. Total tower height = 7 + 7√3 = 7(1+√3) m ≈ 19.12 m
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.
Answer
tan 45° = 75/d₁
⇒ d₁ = 75. tan 30° = 75/d₂
⇒ d₂ = 75√3.
Distance between ships = d₂ – d₁ = 75√3 – 75 = 75(√3–1) m ≈ 54.9 m
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.
Answer
Effective height = 88.2 – 1.2 = 87 m.
At first: tan 60° = 87/d₁
⇒ d₁ = 87/√3 = 29√3.
At second: tan 30° = 87/d₂ → d₂ = 87√3.
Distance = d₂ – d₁ = 87√3 – 29√3 = 58√3 m ≈ 100.46 m
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 is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
Answer
Let height = h. At 30°: d₁ = h/tan 30° = h√3. At 60°: d₂ = h/tan 60° = h/√3.
Distance covered in 6 s = h√3 – h/√3 = 2h/√3. Speed = 2h/(6√3) = h/(3√3).
Time to cover d₂ = (h/√3) / (h/(3√3)) = (h/√3) × (3√3/h) = 3 seconds
Important Extra Questions
Define angle of elevation and angle of depression.
Answer
Angle of elevation: The angle formed between the horizontal line of sight and the line of sight directed upward to an object above the observer. It is measured above the horizontal.
Angle of depression: The angle formed between the horizontal line of sight and the line of sight directed downward to an object below the observer. It is measured below the horizontal. The angle of depression from A to B equals the angle of elevation from B to A (alternate interior angles).
The shadow of a tower standing on 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.
Answer
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 = 2h/√3
⇒ h = 40√3/2 = 20√3 m ≈ 34.64 m