Question

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. The angle of elevation of the balloon from the eyes of the girl at any instant is 60 degree. After some time, the angle of elevation reduces to 30 degree find the distance travelled by the balloon during the interval.

Answer 1

Hi!!friends i have worked it out for u.

Answer 2

Given ,

1. Height of the girl = 1.2m
2. Height of the balloon from the ground = 88.2m
3. angle of elevation from the eye of girl = 60°
4. angle of elevation after some time = 30°

to find :

The distance travelled by the balloon during the interval.

solution :

From ΔBEC,

$\tan(30) = \frac{BE}{CE}$

1/√3= 87/CE

$\frac{1}{ \sqrt{3} } = \frac{87}{CE}$

$CE = 87 \sqrt{3}$

From ΔADC,

$\tan(60) = \frac{AD}{CD }$

$\sqrt{3 } = \frac{87}{CD}$

$CD = \frac{87}{ \sqrt{3} } = 29 \sqrt{3}$

DE = CE – CD

$DE = 87 \sqrt{3} - 29 \sqrt{3} = 58 \sqrt{3}$

Hence, Distance travelled by the balloon = 58√3 m.