# 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 Fig. 9.13). Find the distance travelled by the balloon during the interval.

**Solution:**

Reasoning:

Trigonometric ratio involving AB, BC, OD, OA and angles is tanθ. [Refer AB, BC, OA and OD from the figure.]

Distance travelled by the balloon OB = AB - OA

From the figure, OD = BC, can be calculated as

88.2 m - 1.2 m = 87 m ---(1)

In ΔAOD,

tan 60° = OD/OA

√3 = 87/OA

OA = 87 / √3

= 87 × √3 / √3 × √3

= (87 × √3)/3

= 29√3 m

In ΔABC,

tan 30° = BC/AB

1/√3 = 87 / AB

AB = 87√3

Distance travelled by the balloon, OB = AB - OA

OB = 87√3 - 29√3

= 58√3

Distance travelled by the balloon = 58√3 m.

**Video Solution:**

## 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 Fig. 9.13). Find the distance travelled by the balloon during the interval

### Maths NCERT Solutions Class 10 - Chapter 9 Exercise 9.1 Question 14:

If a 1.2m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2m from the ground and the angle of elevation of the balloon from the eyes of the girl at any instant is 60° and after some time, the angle of elevation reduces to 30°, then the distance travelled by the balloon during the interval is 58√3m.