# Exercise 9.1

**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 30Â° (see figure ).**

**Answer:**

Let AB be the vertical pole Ac be 20 m long rope tied to point C.

In right Î”ABC,

sin 30Â° = AB/AC

â‡’ 1/2 = AB/20

â‡’ AB = 20/2

â‡’ AB = 10

The height of the pole is 10 m.

**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 30Â° 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 be the broken part of the tree.

âˆ´ Total height of the tree = AB+AC

In right Î”ABC,

cos 30Â° = BC/AC

â‡’ âˆš3/2 = 8/AC

â‡’ AC = 16/âˆš3

Also,

tan 30Â° = AB/BC

â‡’ 1/âˆš3 = AB/8

â‡’ AB = 8/âˆš3

Total height of the tree = AB+AC = 16/âˆš3 + 8/âˆš3 = 24/âˆš3

**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 30Â° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60Â° to the ground. What should be the length of the slide in each case?**

**Answer:**

There are two slides of height 1.5 m and 3 m. (Given)

Let AB is 1.5 m and PQ be 3 m slides.

ABC is the slide inclined at 30Â° with length AC and PQR is the slide inclined at

60Â° with length PR.

A/q,

In right Î”ABC,

sin 30Â° = AB/AC

â‡’ 1/2 = 1.5/AC

â‡’ AC = 3m

also,

In right Î”PQR,

sin 60Â° = PQ/PR

â‡’ âˆš3/2 = 3/PR

â‡’ PR = 2âˆš3 m

Hence, length of the slides are 3 m and 2âˆš3 m respectively.

**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 30Â°. Find the height of the tower.**

**Answer:**

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

#### In right Î”ABC,

tan 30Â° = AB/BC

â‡’ 1/âˆš3 = AB/30

â‡’ AB = 10âˆš3

Thus, the height of the tower is 10âˆš3 m.

**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 60Â°. Find the length of the string, assuming that there is no slack in the string.**

**Answer:**

Let BC be the height of the kite from the ground,

AC be the inclined length of the string from the ground and A is the point where string of the kite is tied.

A/q,

In right Î”ABC,

sin 60Â° = BC/AC

â‡’ âˆš3/2 = 60/AC

â‡’ AC = 40âˆš3 m

Thus, the length of the string from the ground is 40âˆš3 m.

**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 30Â° to 60Â° as he walks towards the building. Find the distance he walked towards the building.**

**Answer:**

Let the boy initially standing at point Y with inclination 30Â° and then he approaches the building to

the point X with inclination 60Â°.

âˆ´ XY is the distance he walked towards the building.

also, XY = CD.

Height of the building = AZ = 30 m

AB = AZ – BZ = (30 – 1.5) = 28.5 m

A/q,

In right Î”ABD,

tan 30Â° = AB/BD

â‡’ 1/âˆš3 = 28.5/BD

â‡’ BD = 28.5âˆš3 m

also,

In right Î”ABC,

tan 60Â° = AB/BC

â‡’ âˆš3 = 28.5/BC

â‡’ BC = 28.5/âˆš3 = 28.5âˆš3/3 m

âˆ´ XY = CD = BD – BC = (28.5âˆš3 – 28.5âˆš3/3) = 28.5âˆš3(1-1/3) = 28.5âˆš3 Ã— 2/3 = 57/âˆš3 m.

Thus, the distance boy walked towards the building is 57/âˆš3 m.

**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 45Â° and 60Â° respectively. Find the height of the tower.**

**Answer:**

Let BC be the 20 m high building.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC – BC

In right Î”BCD,

tan 45Â° = BC/CD

â‡’ 1 = 20/CD

â‡’ CD = 20 m

also,

In right Î”ACD,

tan 60Â° = AC/CD

â‡’ âˆš3 = AC/20

â‡’ AC = 20âˆš3 m

Height of transmission tower = AB = AC – BC = (20âˆš3 – 20) m = 20(âˆš3 – 1) m.

**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 60Â° and from the same point the angle of elevation of the top of the pedestal is 45Â°. Find the height of the pedestal.**

**Answer:**