#### Page No 203:

#### Question 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 °.

#### Answer:

It can be observed from the figure that AB is the pole.

In ΔABC,

Therefore, the height of the pole is 10 m.

#### Page No 203:

#### Question 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.

#### Answer:

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part is making 30° with the ground.

In ,

Height of tree = + BC

Hence, the height of the tree is.

[[VIDEO:14279]]

#### Page No 203:

#### Question 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, where as for the elder children she wants to have a steep side at a height of 3 m, and inclined at an angle of 60 ° to the ground. What should be the length of the slide in each case?

#### Answer:

It can be observed that AC and PR are the slides for younger and elder children respectively.

In ΔABC,

In ΔPQR,

Therefore, the lengths of these slides are 3 m and .

#### Page No 204:

#### Question 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.

#### Answer:

Let AB be the tower and the angle of elevation from point C (on ground) is

30°.

In ΔABC,

Therefore, the height of the tower is.

#### Page No 204:

#### Question 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.

#### Answer:

Let K be the kite and the string is tied to point P on the ground.

In ΔKLP,

Hence, the length of the string is.

#### Page No 204:

#### Question 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.

#### Answer:

Let the boy was standing at point S initially. He walked towards the building and reached at point T.

It can be observed that

PR = PQ − RQ

= (30 − 1.5) m = 28.5 m =

In ΔPAR,

In ΔPRB,

ST = AB

Hence, he walked towards the building.

[[VIDEO:14280]]

#### Page No 204:

#### Question 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.

#### Answer:

Let BC be the building, AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,

In ΔACD,

Therefore, the height of the transmission tower is m.

#### Page No 204:

#### Question 8:

A statue, 1.6 m tall, stands on a top of pedestal, from a point on the ground, the angle of elevation of the top of 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.

#### Answer:

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,

In ΔACD,

Therefore, the height of the pedestal is 0.8m.

#### Page No 204:

#### Question 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.

#### Answer:

Let AB be the building and CD be the tower.

In ΔCDB,

In ΔABD,

Therefore, the height of the building is.

[[VIDEO:14281]]

#### Page No 204:

#### Question 10:

Two poles of equal heights are standing opposite each other an 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 30º, respectively. Find the height of poles and the distance of the point from the poles.

#### Answer:

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ΔABO,

In ΔCDO,

Since the poles are of equal heights,

CD = AB

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles isand the point is 20 m and 60 m far from these poles.

#### Page No 204:

#### Question 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 the tower is 30°. Find the height of the tower and the width of the canal.

#### Answer:

In ΔABC,

In ΔABD,

Therefore, the height of the tower is m and the width of the canal is

10 m.

#### Page No 204:

#### Question 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.

#### Answer:

Let AB be a building and CD be a cable tower.

In ΔABD,

In ΔACE,

AE = BD = 7 m

Therefore, the height of the cable tower is.

#### Page No 204:

#### Question 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.

#### Answer:

Let AB be the lighthouse and the two ships be at point C and D respectively.

In ΔABC,

In ΔABD,

Therefore, the distance between the two ships ism.

[[VIDEO:14282]]

#### Page No 205:

#### Question 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°. Find the distance travelled by the balloon during the interval.

#### Answer:

Let the initial position A of balloon change to B after some time and CD be the girl.

In ΔACE,

In ΔBCG,

Distance travelled by balloon = EG = CG − CE

[[VIDEO:14283]]

#### Page No 205:

#### Question 15:

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as 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.

#### Answer:

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ΔADB,

In ΔABC,

Time taken by the car to travel distance DC= 6 seconds

Time taken by the car to travel distance DB

#### Page No 205:

#### Question 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.

#### Answer:

Let AQ be the tower and R, S are the points 4m, 9m away from the base of the tower respectively.

The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ΔAQR,

In ΔAQS,

On multiplying equations (*i*) and (*ii*), we obtain

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

View NCERT Solutions for all chapters of Class 10