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

Let CD be the tower.
∴ CD = 50 m
Now, in right ΔABC,


In right ΔDCA,


From (1) and (2), we get

Thus, the height of the building

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

CD = h metres
AP = x m
CP = (80 - x) m
Now, in right ΔAPB,
we have


Again in right ΔCPD, 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.

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.

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


In right ΔABD, 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.

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.

Let CD be the cable tower.
∴ In right ΔDAE, we have

...(1)
Again, in right ΔABC, we have

...(2)
From (1) and (2), we get

∴ CD = CE + ED
        
        = 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



Again, in right ΔABD, 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.

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.

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


In right ΔPDC, we have


Now, BD = CD - BC




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

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.

In right ΔABD, we have

...(1)
In right ΔABC, we have

...(2)
From (1) and (2), we get


Now, CD = AC - AD = 3AD - AD = 2AD
Since the distance 2AD 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.

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.

Let AB = h metres.
∴ In right Δ ABC, we have

...(1)
In right ΔABD, 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.