Class 10 - Chapter 9: Some Applications of Trigonometry

Exercise 9.1

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°.

In right △ABC,

AC = 20 m (rope)

∠C = 30°

AB = height of pole

Using:

sin 30° = AB / AC

1/2 = AB / 20

AB = 20 × 1/2

AB = 10 m

Height of the pole = 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 and the point where the top touches the ground is 8 m. Find the height of the tree.

Let AB be the unbroken part of the tree and AC be the broken part.

BC = 8 m

∠C = 30°

Using:

tan 30° = AB / BC

1/√3 = AB / 8

AB = 8/√3

AB = 8√3/3 m

Also,

sin 30° = AB / AC

1/2 = AB / AC

AC = 2AB

AC = 16√3/3 m

Height of tree = AB + AC

= 8√3/3 + 16√3/3

= 24√3/3

= 8√3 m

Height of the tree = 8√3 m ≈ 13.86 m

Q3. A contractor plans to install two slides for children. Find the length of each slide.

(i) Slide for children below 5 years

Height = 1.5 m

Angle with ground = 30°

Let length of slide = L

sin 30° = 1.5 / L

1/2 = 1.5 / L

L = 3 m

(ii) Slide for elder children

Height = 3 m

Angle with ground = 60°

sin 60° = 3 / L

√3/2 = 3 / L

L = 6/√3

L = 2√3 m

≈ 3.46 m

Lengths of slides:

• First slide = 3 m
• Second slide = 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.

Distance from tower = 30 m

Angle of elevation = 30°

Let height of tower = h

tan 30° = h / 30

1/√3 = h / 30

h = 30/√3

h = 10√3 m

≈ 17.32 m

Height of tower = 10√3 m

Q5. A kite is flying at a height of 60 m above the ground. The string attached to the kite makes an angle of 60° with the ground. Find the length of the string.

Height of kite = 60 m

Angle with ground = 60°

Let length of string = L

sin 60° = 60 / L

√3/2 = 60 / L

L = 120/√3

L = 40√3 m

≈ 69.28 m

Length of string = 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.

Height of building = 30 m

Height of boy's eyes = 1.5 m

Vertical height observed:

= 30 − 1.5

= 28.5 m

Let initial distance = x m

tan 30° = 28.5 / x

1/√3 = 28.5 / x

x = 28.5√3

Let final distance = y m

tan 60° = 28.5 / y

√3 = 28.5 / y

y = 28.5/√3

y = 9.5√3

Distance walked

= x − y

= 28.5√3 − 9.5√3

= 19√3 m

≈ 32.91 m

Distance walked = 19√3 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.

Let height of tower = h m

Let distance of observation point from building = x m

For the bottom of tower:

tan 45° = 20 / x

1 = 20 / x

x = 20 m

For the top of tower:

tan 60° = (20 + h)/20

√3 = (20 + h)/20

20√3 = 20 + h

h = 20√3 − 20

h = 20(√3 − 1)

≈ 14.64 m

Height of the transmission tower = 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.

Let the height of the pedestal be h m and the distance of the observation point from the pedestal be x m.

For the top of the pedestal:

tan 45° = h/x

1 = h/x

x = h

For the top of the statue:

tan 60° = (h + 1.6)/x

√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)

≈ 2.19 m

Height of the pedestal = 0.8(√3 + 1) m ≈ 2.19 m

Q9. The angle of elevation of the top of a building from the foot of a 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.

Let the height of the building be h m and the distance between the tower and the building be x m.

From the foot of the building:

tan 60° = 50/x

√3 = 50/x

x = 50/√3

From the foot of the tower:

tan 30° = h/x

1/√3 = h/(50/√3)

h = 50/3

≈ 16.67 m

Height of the building = 50/3 m ≈ 16.67 m

Q10. Two poles of equal heights are standing opposite each other on either side of a road, which is 80 m wide. From a point between them on the road, the angles of elevation of the tops of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Let the distances of the point from the poles be x m and (80 − x) m.

Let the common height of the poles be h m.

For the first pole:

tan 60° = h/x

√3 = h/x

h = x√3

For the second pole:

tan 30° = h/(80 − x)

1/√3 = h/(80 − x)

h = (80 − x)/√3

Equating the two values of h:

x√3 = (80 − x)/√3

3x = 80 − x

4x = 80

x = 20

Distance from the second pole:

80 − 20 = 60 m

Height:

h = 20√3 m

≈ 34.64 m

Height of each pole = 20√3 m

Distances from the poles = 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.

Let:

Height of tower = h m

Width of canal = BC = x m

From point C:

tan 60° = h/x

√3 = h/x

h = x√3

From point D:

DB = x + 20

tan 30° = h/(x + 20)

1/√3 = h/(x + 20)

h = (x + 20)/√3

Equating:

x√3 = (x + 20)/√3

3x = x + 20

2x = 20

x = 10

Therefore,

h = 10√3 m

≈ 17.32 m

Width of canal = 10 m

Height of tower = 10√3 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.

Let AB be the building of height 7 m and CD be the tower.

Let the horizontal distance between them be x m.

From the angle of depression:

tan 45° = 7/x

1 = 7/x

x = 7 m

Let the height of the tower be h m.

Using the angle of elevation:

tan 60° = (h − 7)/7

√3 = (h − 7)/7

h − 7 = 7√3

h = 7(√3 + 1)

≈ 19.12 m

Height of the tower = 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.

Height of lighthouse = 75 m

Let the distances of the ships from the lighthouse be x and y.

For the nearer ship (45°)

tan 45° = 75/x

1 = 75/x

x = 75 m

For the farther ship (30°)

tan 30° = 75/y

1/√3 = 75/y

y = 75√3 m

Distance between the ships:

= y − x

= 75√3 − 75

= 75(√3 − 1)

≈ 54.90 m

Distance between the ships = 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.

Height of the balloon above the girl's eyes:

= 88.2 − 1.2

= 87 m

Let the horizontal distance of the balloon from the girl initially be x m.

When the angle of elevation is 60°:

tan 60° = 87/x

√3 = 87/x

x = 87/√3

x = 29√3 m

Let the horizontal distance after some time be y m.

When the angle of elevation is 30°:

tan 30° = 87/y

1/√3 = 87/y

y = 87√3 m

Distance travelled by the balloon:

= y − x

= 87√3 − 29√3

= 58√3 m

≈ 100.46 m

Distance travelled by the balloon = 58√3 m ≈ 100.46 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.

Let:

Height of the tower = h m

Initial distance of the car from the foot of the tower = x m

Distance of the car after 6 seconds = y m

At the first position

Angle of depression = 30°

Therefore, angle of elevation = 30°

tan 30° = h/x

1/√3 = h/x

x = h√3

After 6 seconds

Angle of depression = 60°

Therefore, angle of elevation = 60°

tan 60° = h/y

√3 = h/y

y = h/√3

Distance travelled in 6 seconds

= x − y

= h√3 − h/√3

= (3h − h)/√3

= 2h/√3

Speed of the car:

= (2h/√3)/6

= h/(3√3) m/s

Time to reach the tower from the second position

Time = Distance / Speed

= (h/√3) ÷ (h/(3√3))

= (h/√3) × (3√3/h)

= 3 seconds

Time taken by the car to reach the foot of the tower from the second position = 3 seconds.