Applications of Trigonometry - NCERT Questions

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 by the rope with the ground level is 30° (see figure).

**Soln. :** In right Δ*ABC*,

[∵* AC *= 20 m and sin 30° ]

Thus, the required height of the pole is 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 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.

**Soln. :** Let tree is broken at *A* and its top is touching the ground at *B*.

Now, in right Δ*AOB*, we have

m

m

Now, height of the tree *OP* = *OA* +* AB*

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 angle of 60° to the ground. What should be the length of the slide in each case ?

SOLUTION:
**Soln. :** In the figure, *DE* is the slide for younger children whereas *AC* is the slide for elder children.

In right Δ*ABC*, *AB* = 3 m

Again in right Δ*BDE*,

⇒ * DE* = 2 × 1.5 m ⇒ *DE* = 3 m

Thus, the lengths of slides are 3 m and

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:
**Soln. :** In right Δ*ABC*, *AB* = height of the tower and point *C* is 30 m away from the foot of the tower,

∴ *AC* = 30 m

Now,

[∵ tan 30° = ]

Thus, the required height of the tower is .

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.

**Soln. :** Let *OB* = Length of the string
* AB* = 60 m = Height of the kite.

In the right Δ*AOB*,

Thus, length of the string is .

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:
**Soln. :** Here, *OA* is the building.

In right Δ*ABD*,

[∵* AD *= 30 m - 1.5 m = 28.5 m]

Also, in right Δ*ACD*,

Now,

Now,

Thus the distance walked by the boy towards the building is

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.

**Soln. :** Let the height of the building be *BC*

∴* BC* = 20 m

and height of the tower be *CD*.

Let the point *A* be at a distance *y* metres from the foot of the building.

Now, in right Δ*ABC*,

Now, in right Δ *ABD*,

⇒ *x* = 20 × 0.732 = 14.64

Thus, the height of the tower is 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.

SOLUTION:
**Soln. :** In the figure, *DC* represents the statue and *BC* represents the pedestal.

Now, in right Δ*ABC*, we have

Now in right Δ*ABD*,

we have

Thus, the height of the pedestal is

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:
**Soln. :** In the figure, let height of the building = *AB* =* h* m

Let *CD* be the tower.

∴ *CD* = 50 m

Now, in right Δ*ABC*,

In right Δ*DCA*,

From (1) and (2), we get

Thus, the height of the building

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:
**Soln. :** *AB* = *h* metres

*CD *= *h* metres
*AP* = *x* m
*CP* = (80 - *x*) m

Now, in right Δ*APB*,

we have

Again in right Δ*CPD*, we have

...(2)

From (1) and (2), we get

∴ *CP* = 80 -* x* = 80 - 20 = 60 m

Now, from (1), we have

Thus, the required point is 20 m away from the first pole and 60 m away from the second pole.

Height of each pole = 34.64 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 of tower is 30° (see figure). Find the height of the tower and the width of the canal.

**Soln. :** Let the TV tower be *AB* = *h *m.

Let the point '*C*' be such that* BC* = *x* and *CD* = 20 m.

Now, in right Δ*ABC*, we have

In right Δ*ABD*, we have

From (1) and (2), we get

Now, from (1), we get

Thus, the height of the tower = 17.32 m.

Also width of the canal = 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.

SOLUTION:
**Soln. :** In the figure, let *AB* be the height of the building.

∴* AB* = 7 metres.

Let *CD *be the cable tower.

∴ In right Δ*DAE*, we have

...(1)

Again, in right Δ*ABC*, we have

...(2)

From (1) and (2), we get

∴ *CD* = *CE* + *ED*

= 7(1 + 1.732)m = 7(2.732)m = 19.124 m

As observed from the top of a 75 m high light house 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 light house, find the distance between the two ships.

SOLUTION:
**Soln. :** In the figure, let* AB *represent the light house.

∴ *AB* = 75 m.

Let the two ships be* C* and *D* such that angle of depression from *A* are 45° and 30° respectively.

Now, in right Δ*ABC*,

we have

Again, in right Δ*ABD*, we have

Since the distance between the two ships = *CD*

=75[1.732 -1] = 75 × 0.732 = 54.9

Thus, the required distance between the ships is 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° (see figure). Find the distance travelled by the balloon during the interval.

**Soln. :** In the figure, let *C* be the position of the observer (the girl).
*A* and *P* are two positions of the balloon.
*CD* is the horizontal line from the eyes of the observer (girl).

Here* PD* = *AB *= 88.2 m - 1.2 m = 87 m

In right Δ*ABC*, we have

In right Δ*PDC*, we have

Now, *BD *= *CD* - *BC*

Thus, the required distance between the two positions of the balloon

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:
**Soln. :** In the figure, let* AB* be the height of the tower and *C*, *D* be the two positions of the car.

In right Δ*ABD*, we have

...(1)

In right Δ*ABC*, we have

...(2)

From (1) and (2), we get

Now, *CD* = *AC *- *AD* = 3*AD *- *AD *= 2*AD*

Since the distance 2*AD* is covered in 6 seconds,

∴ The distance *AD* will be covered in *i.e*., 3 seconds,

Thus, the time taken by the car to reach the tower from *D* is 3 seconds.

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:
**Soln. :** Let the tower be represented by *AB* in the figure.

Let *AB *=* h* metres.

∴ In right Δ* ABC*, we have

...(1)

In right Δ*ABD*, we have

...(2)

Multiplying (1) and (2), we get

[∵ tan θ × cot θ = 1]

⇒ *h* = ± 6 m ∴ *h* = 6 m

[∵ Height is positive only]

Thus, the height of the tower is 6 m.