Exercise 9.1 Solutions
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°.
Given: Length of rope = 20 m, angle with ground = 30°
To find: Height of the pole
Solution:
sin 30° = Height of pole / Length of rope
1/2 = Height / 20
Height = 20 × 1/2 = 10 m
Answer: The height of the pole is 10 m.
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.
Given: Distance from tree foot to top contact point = 8 m, angle = 30°
To find: Height of the tree
Solution:
Let the height of the standing part be h and the length of the broken part be l.
tan 30° = h / 8
1/√3 = h / 8
h = 8/√3 m
cos 30° = 8 / l
√3/2 = 8 / l
l = 16/√3 m
Total height = h + l = 8/√3 + 16/√3 = 24/√3 = 8√3 m
Answer: The height of the tree was 8√3 m.
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 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?
For younger children:
Height = 1.5 m, angle = 30°
sin 30° = Height / Length of slide
1/2 = 1.5 / Length
Length = 1.5 × 2 = 3 m
For elder children:
Height = 3 m, angle = 60°
sin 60° = Height / Length of slide
√3/2 = 3 / Length
Length = 3 × 2/√3 = 6/√3 = 2√3 m
Answer: The length of the slide for younger children is 3 m and for elder children is 2√3 m.
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.
Given: Distance from tower = 30 m, angle of elevation = 30°
To find: Height of tower
Solution:
tan 30° = Height / Distance
1/√3 = Height / 30
Height = 30/√3 = 10√3 m
Answer: The height of the tower is 10√3 m.
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.
Given: Height = 60 m, angle = 60°
To find: Length of string
Solution:
sin 60° = Height / Length of string
√3/2 = 60 / Length
Length = 60 × 2/√3 = 120/√3 = 40√3 m
Answer: The length of the string is 40√3 m.
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.
Given: Height of boy = 1.5 m, height of building = 30 m
Initial angle of elevation = 30°, final angle of elevation = 60°
To find: Distance walked
Solution:
Effective height from boy's eyes = 30 - 1.5 = 28.5 m
Let initial distance be d₁ and final distance be d₂
tan 30° = 28.5 / d₁
1/√3 = 28.5 / d₁
d₁ = 28.5√3 m
tan 60° = 28.5 / d₂
√3 = 28.5 / d₂
d₂ = 28.5/√3 m
Distance walked = d₁ - d₂ = 28.5√3 - 28.5/√3
= 28.5(√3 - 1/√3) = 28.5(3-1)/√3 = 28.5 × 2/√3 = 57/√3 = 19√3 m
Answer: The boy walked 19√3 m towards the building.
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.
Given: Height of building = 20 m
Angle of elevation to bottom of tower = 45°
Angle of elevation to top of tower = 60°
To find: Height of tower
Solution:
Let the distance from the point to the building be d
tan 45° = 20 / d
1 = 20 / d
d = 20 m
Let the height of the tower be h
tan 60° = (20 + h) / d
√3 = (20 + h) / 20
20 + h = 20√3
h = 20√3 - 20 = 20(√3 - 1) m
Answer: The height of the tower is 20(√3 - 1) m.
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.
Given: Height of statue = 1.6 m
Angle of elevation to top of statue = 60°
Angle of elevation to top of pedestal = 45°
To find: Height of pedestal
Solution:
Let the height of pedestal be h and distance from point be d
tan 45° = h / d
1 = h / d
d = h
tan 60° = (h + 1.6) / d
√3 = (h + 1.6) / h
√3h = h + 1.6
√3h - h = 1.6
h(√3 - 1) = 1.6
h = 1.6 / (√3 - 1)
Rationalizing: h = 1.6(√3 + 1) / (3 - 1) = 1.6(√3 + 1) / 2 = 0.8(√3 + 1) m
Answer: The height of the pedestal is 0.8(√3 + 1) m.
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.
Given: Height of tower = 50 m
Angle of elevation from tower foot to building top = 30°
Angle of elevation from building foot to tower top = 60°
To find: Height of building
Solution:
Let the distance between tower and building be d
tan 60° = 50 / d
√3 = 50 / d
d = 50/√3 m
Let the height of building be h
tan 30° = h / d
1/√3 = h / (50/√3)
1/√3 = h√3 / 50
h = 50 / 3 = 50/3 m
Answer: The height of the building is 50/3 m.
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.
Given: Width of road = 80 m
Angles of elevation: 60° and 30°
To find: Height of poles and distances from point
Solution:
Let the height of poles be h
Let distances from point to poles be x and (80 - x)
tan 60° = h / x
√3 = h / x
h = x√3
tan 30° = h / (80 - x)
1/√3 = h / (80 - x)
Substituting h = x√3:
1/√3 = x√3 / (80 - x)
80 - x = 3x
80 = 4x
x = 20 m
h = 20√3 m
Answer: The height of poles is 20√3 m. The distances from the point to the poles are 20 m and 60 m.
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 going this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.
Given: Distance between two points = 20 m
Angles of elevation: 60° and 30°
To find: Height of tower and width of canal
Solution:
Let the height of tower be h and width of canal be d
tan 60° = h / d
√3 = h / d
h = d√3
tan 30° = h / (d + 20)
1/√3 = h / (d + 20)
Substituting h = d√3:
1/√3 = d√3 / (d + 20)
d + 20 = 3d
20 = 2d
d = 10 m
h = 10√3 m
Answer: The height of the tower is 10√3 m and the width of the canal is 10 m.
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.
Given: Height of building = 7 m
Angle of elevation to tower top = 60°
Angle of depression to tower foot = 45°
To find: Height of tower
Solution:
Let the distance between building and tower be d
tan 45° = 7 / d
1 = 7 / d
d = 7 m
Let the height of tower above building be h
tan 60° = h / d
√3 = h / 7
h = 7√3 m
Total height of tower = 7 + 7√3 = 7(1 + √3) m
Answer: The height of the tower is 7(1 + √3) m.
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.
Given: Height of lighthouse = 75 m
Angles of depression: 30° and 45°
To find: Distance between two ships
Solution:
Let the distances from lighthouse to ships be d₁ and d₂
tan 45° = 75 / d₁
1 = 75 / d₁
d₁ = 75 m
tan 30° = 75 / d₂
1/√3 = 75 / d₂
d₂ = 75√3 m
Distance between ships = d₂ - d₁ = 75√3 - 75 = 75(√3 - 1) m
Answer: The distance between the two ships is 75(√3 - 1) m.
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 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. 9.13). Find the distance travelled by the balloon during the interval.
Given: Height of girl = 1.2 m, height of balloon = 88.2 m
Initial angle of elevation = 60°, final angle of elevation = 30°
To find: Distance travelled by balloon
Solution:
Effective height from girl's eyes = 88.2 - 1.2 = 87 m
Let initial horizontal distance be d₁ and final be d₂
tan 60° = 87 / d₁
√3 = 87 / d₁
d₁ = 87/√3 m
tan 30° = 87 / d₂
1/√3 = 87 / d₂
d₂ = 87√3 m
Distance travelled = d₂ - d₁ = 87√3 - 87/√3
= 87(√3 - 1/√3) = 87(3-1)/√3 = 87 × 2/√3 = 174/√3 = 58√3 m
Answer: The balloon travelled 58√3 m during the interval.
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.
Given: Initial angle of depression = 30°
Final angle of depression (after 6 seconds) = 60°
To find: Time taken to reach foot of tower
Solution:
Let the height of tower be h
Let initial distance be d₁ and final distance be d₂
tan 30° = h / d₁
1/√3 = h / d₁
d₁ = h√3
tan 60° = h / d₂
√3 = h / d₂
d₂ = h/√3
Distance covered in 6 seconds = d₁ - d₂ = h√3 - h/√3 = h(√3 - 1/√3) = h(3-1)/√3 = 2h/√3
Speed of car = Distance / Time = (2h/√3) / 6 = h/(3√3)
Remaining distance to foot of tower = d₂ = h/√3
Time = Distance / Speed = (h/√3) / (h/(3√3)) = 3 seconds
Answer: The car will take 3 seconds to reach the foot of the tower.