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Applications of Trigonometry - NCERT Questions
Q 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 by the rope with the ground level is 30° (see figure).

SOLUTION:

Soln. : In right ΔABC,


         [∵ AC = 20 m and sin 30° ]

Thus, the required height of the pole is 10 m.

Q 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.

SOLUTION:

Soln. : Let tree is broken at A and its top is touching the ground at B.
Now, in right ΔAOB, we have


m

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

Q 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 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 ΔABC, AB = 3 m


         Again in right ΔBDE,

DE = 2 × 1.5 m ⇒ DE = 3 m
Thus, the lengths of slides are 3 m and

Q 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.

SOLUTION:

Soln. : In right ΔABC, 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 .

Q 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.

SOLUTION:

Soln. : Let OB = Length of the string
AB = 60 m = Height of the kite.
In the right ΔAOB,



Thus, length of the string is .

Q 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.

SOLUTION:

Soln. : Here, OA is the building.

In right ΔABD,


         [∵ AD = 30 m - 1.5 m = 28.5 m]
Also, in right ΔACD,

Now,
Now,



Thus the distance walked by the boy towards the building is

Q 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.

SOLUTION:

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 ΔABC,


Now, in right Δ ABD,




x = 20 × 0.732 = 14.64
Thus, the height of the tower is 14.64 m.

Q 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.

SOLUTION:

Soln. : In the figure, DC represents the statue and BC represents the pedestal.
Now, in right ΔABC, we have



Now in right ΔABD,
we have





Thus, the height of the pedestal is

Q 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.

SOLUTION:

Soln. : In the figure, let height of the building = AB = h m
ncert
Let CD be the tower.
CD = 50 m
Now, in right ΔABC,
ncert
ncert
In right ΔDCA,
ncert
ncert
From (1) and (2), we get
ncert
Thus, the height of the buildingncert

Q 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.

SOLUTION:

Soln. : AB = h metres
ncert
CD = h metres
AP = x m
CP = (80 - x) m
Now, in right ΔAPB,
we have
ncert
ncert
Again in right ΔCPD, we have
ncert
ncert ncert ...(2)
From (1) and (2), we get
ncert
ncert
ncert
CP = 80 - x = 80 - 20 = 60 m
Now, from (1), we have
ncert
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.

Q 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 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.
ncert

SOLUTION:

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 ΔABC, we have
ncert
ncert
In right ΔABD, we have
ncert
ncert
From (1) and (2), we get
ncert
ncert
Now, from (1), we get
ncert
Thus, the height of the tower = 17.32 m.
Also width of the canal = 10 m.

Q 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.

SOLUTION:

Soln. : In the figure, let AB be the height of the building.
AB = 7 metres.
ncert
Let CD be the cable tower.
∴ In right ΔDAE, we have
ncert
ncert ...(1)
Again, in right ΔABC, we have
ncert
ncert ...(2)
From (1) and (2), we get
ncert
CD = CE + ED
        ncert
        = 7(1 + 1.732)m = 7(2.732)m = 19.124 m

Q 13.

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 ΔABC,
we have
ncert
ncert
ncert
Again, in right ΔABD, we have
ncert
Since the distance between the two ships = CD
ncert
=75[1.732 -1] = 75 × 0.732 = 54.9
Thus, the required distance between the ships is 54.9 m.

Q 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 figure). Find the distance travelled by the balloon during the interval.
ncert

SOLUTION:

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 ΔABC, we have
ncert
ncert
In right ΔPDC, we have
ncert
ncert
Now, BD = CD - BC
ncert
ncert
ncert
ncert
Thus, the required distance between the two positions of the balloonncert

Q 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.

SOLUTION:

Soln. : In the figure, let AB be the height of the tower and C, D be the two positions of the car.
ncert
In right ΔABD, we have
ncert
ncert ...(1)
In right ΔABC, we have
ncert
ncert ...(2)
From (1) and (2), we get
ncert
ncert
Now, CD = AC - AD = 3AD - AD = 2AD
Since the distance 2AD is covered in 6 seconds,
∴ The distance AD will be covered in ncerti.e., 3 seconds,
Thus, the time taken by the car to reach the tower from D is 3 seconds.

Q 16.

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.
ncert
Let AB = h metres.
∴ In right Δ ABC, we have
ncert
ncert ...(1)
In right ΔABD, we have
ncert
ncert ...(2)
Multiplying (1) and (2), we get
ncert [∵ tan θ × cot θ = 1]
ncert
h = ± 6 m ∴ h = 6 m
             [∵ Height is positive only]
Thus, the height of the tower is 6 m.