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°.

Solution:

Let AB be the vertical pole and AC be the 20 m long rope attached at point C on the ground.
Given: Length of rope (AC) = 20 m, Angle of elevation ∠ACB = 30°.

In right-angled triangle ABC:
sin 30° = AB / AC
1/2 = AB / 20
AB = 20 / 2 = 10 m

Answer: The height of the pole is 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.

Solution:

Let AB be the unbroken lower part and AC be the broken upper part of the tree. The total height of the tree is (AB + AC).
Given: BC = 8 m, ∠ACB = 30°.

In right-angled triangle ABC:
tan 30° = AB / BC
1/√3 = AB / 8 => AB = 8/√3 m

Also, cos 30° = BC / AC
√3/2 = 8 / AC => AC = 16/√3 m

Total height of the tree = AB + AC = 8/√3 + 16/√3 = 24/√3 = 8√3 m.

Answer: The height of the tree is 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?

Solution:

Case 1: For children below 5 years
Height (h1) = 1.5 m, Angle = 30°
sin 30° = 1.5 / L1
1/2 = 1.5 / L1 => L1 = 3 m

Case 2: For elder children
Height (h2) = 3 m, Angle = 60°
sin 60° = 3 / L2
√3/2 = 3 / L2 => L2 = 6/√3 = 2√3 m

Answer: Length of slide for younger children is 3 m and for elder children is 2√3 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.

Solution:

Let h be the height of the tower.
Distance from foot = 30 m, Angle = 30°.

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

Answer: The height of the tower is 10√3 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 string.

Solution:

Let L be the length of the string.
Height = 60 m, Angle = 60°.

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

Answer: The length of the string is 40√3 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.

Solution:

Effective height of the building above boy's eye level = 30 - 1.5 = 28.5 m.
Let initial distance from building be x + d and final distance be x, where d is the distance walked.

In 1st position (60°):
tan 60° = 28.5 / x => √3 = 28.5 / x => x = 28.5 / √3

In 2nd position (30°):
tan 30° = 28.5 / (x + d)
1/√3 = 28.5 / (28.5/√3 + d)
28.5/√3 + d = 28.5√3
d = 28.5√3 - 28.5/√3 = 28.5(3 - 1)/√3 = 57/√3 = 19√3 m

Answer: The boy walked 19√3 m towards the building.

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.

Solution:

Let height of building = 20 m, height of tower = h, and distance of point from building = x.

For bottom of tower (45°):
tan 45° = 20 / x => 1 = 20 / x => x = 20 m

For top of tower (60°):
tan 60° = (20 + h) / x
√3 = (20 + h) / 20
20 + h = 20√3 => h = 20(√3 - 1) m

Answer: The height of the transmission tower is 20(√3 - 1) 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.

Solution:

Let height of pedestal = h.
For pedestal (45°): tan 45° = h / x => x = h.

For top of statue (60°):
tan 60° = (h + 1.6) / x
√3 = (h + 1.6) / h
h√3 = h + 1.6
h(√3 - 1) = 1.6
h = 1.6 / (√3 - 1) = [1.6(√3 + 1)] / 2 = 0.8(√3 + 1) m

Answer: The height of the pedestal is 0.8(√3 + 1) 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.

Solution:

Let height of building = h, height of tower = 50 m, distance between them = x.

From foot of building to top of tower (60°):
tan 60° = 50 / x => x = 50 / √3

From foot of tower to top of building (30°):
tan 30° = h / x
1/√3 = h / (50 / √3)
h = 50 / 3 = 16.67 m

Answer: The height of the building is 16(2/3) m or 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.

Solution:

Let height of poles = h. Let distance of point from 1st pole = x, then distance from 2nd pole = 80 - x.

For 1st pole (60°): tan 60° = h / x => h = x√3
For 2nd pole (30°): tan 30° = h / (80 - x) => h = (80 - x) / √3

Equating both:
x√3 = (80 - x) / √3
3x = 80 - x => 4x = 80 => x = 20 m

Height of poles h = 20√3 m.
Distances from poles are 20 m and 60 m.

Answer: Height of poles is 20√3 m and distances are 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.

Solution:

Let height of tower = h, width of canal = x.

tan 60° = h / x => h = x√3
tan 30° = h / (x + 20) => 1/√3 = (x√3) / (x + 20)
x + 20 = 3x => 2x = 20 => x = 10 m

Height h = 10√3 m.

Answer: Height of tower is 10√3 m and width of canal is 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.

Solution:

Height of building = 7 m. Let total height of cable tower = H, distance between building and tower = x.

Angle of depression to foot (45°):
tan 45° = 7 / x => x = 7 m

Height of tower above building level = H - 7.
tan 60° = (H - 7) / 7
√3 = (H - 7) / 7 => H - 7 = 7√3
H = 7(√3 + 1) m

Answer: Height of the cable tower is 7(√3 + 1) 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.

Solution:

Height of lighthouse = 75 m.
For nearer ship (45°): tan 45° = 75 / x => x = 75 m.
For farther ship (30°): tan 30° = 75 / (x + d) => 1/√3 = 75 / (75 + d)
75 + d = 75√3 => d = 75(√3 - 1) m.

Answer: Distance between ships is 75(√3 - 1) 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.

Solution:

Height above eye level = 88.2 - 1.2 = 87 m.
At 60° position: tan 60° = 87 / x1 => x1 = 87 / √3 = 29√3 m.
At 30° position: tan 30° = 87 / x2 => 1/√3 = 87 / x2 => x2 = 87√3 m.

Distance travelled = x2 - x1 = 87√3 - 29√3 = 58√3 m.

Answer: Distance travelled by the balloon is 58√3 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.

Solution:

Let height of tower be h and speed of car be v.
At 30°: Distance d1 = h / tan 30° = h√3
At 60°: Distance d2 = h / tan 60° = h / √3

Distance covered in 6 s = d1 - d2 = h√3 - h/√3 = 2h / √3.
Speed v = Distance / Time = (2h / √3) / 6 = h / (3√3).

Time to cover remaining distance d2 = d2 / v = (h / √3) / (h / 3√3) = 3 seconds.

Answer: Time taken by the car to reach the foot of the tower is 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.

Solution:

Let height of tower = h. Let angles be θ and (90° - θ).

tan θ = h / 4
tan(90° - θ) = cot θ = h / 9

Multiplying both equations:
tan θ × cot θ = (h / 4) × (h / 9)
1 = h² / 36
h² = 36 => h = 6 m.

Answer: Hence proved, the height of the tower is 6 m.