# 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 30° (see figure ).**

**Answer:**

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

In right ΔABC,

sin 30° = AB/AC

⇒ 1/2 = AB/20

⇒ AB = 20/2

⇒ 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 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 be the broken part of the tree.

∴ Total height of the tree = AB+AC

In right ΔABC,

cos 30° = BC/AC

⇒ √3/2 = 8/AC

⇒ AC = 16/√3

Also,

tan 30° = AB/BC

⇒ 1/√3 = AB/8

⇒ AB = 8/√3

Total height of the tree = AB+AC = 16/√3 + 8/√3 = 24/√3

**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 an angle of 60° 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 30° with length AC and PQR is the slide inclined at

60° with length PR.

A/q,

In right ΔABC,

sin 30° = AB/AC

⇒ 1/2 = 1.5/AC

⇒ AC = 3m

also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3 m

Hence, length of the slides are 3 m and 2√3 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 30°. 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 ΔABC,

tan 30° = AB/BC

⇒ 1/√3 = AB/30

⇒ AB = 10√3

Thus, the height of the tower is 10√3 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 60°. 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 ΔABC,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Thus, the length of the string from the ground is 40√3 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 30° to 60° 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 30° and then he approaches the building to

the point X with inclination 60°.

∴ 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 ΔABD,

tan 30° = AB/BD

⇒ 1/√3 = 28.5/BD

⇒ BD = 28.5√3 m

also,

In right ΔABC,

tan 60° = AB/BC

⇒ √3 = 28.5/BC

⇒ BC = 28.5/√3 = 28.5√3/3 m

∴ XY = CD = BD – BC = (28.5√3 – 28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 m.

Thus, the distance boy walked towards the building is 57/√3 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 45° and 60° 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 ΔBCD,

tan 45° = BC/CD

⇒ 1 = 20/CD

⇒ CD = 20 m

also,

In right ΔACD,

tan 60° = AC/CD

⇒ √3 = AC/20

⇒ AC = 20√3 m

Height of transmission tower = AB = AC – BC = (20√3 – 20) m = 20(√3 – 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 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:**