Question

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.

Answer 1

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

In ∆ACE,

$\frac{AE}{ CE} = tan 60^{\degree}$

$\frac{AF - EF}{ CE} = tan 60^{\degree}$

$\frac{88.2 - 1.2}{ CE} = \sqrt3$

$\frac{87}{ CE} = \sqrt3$

⇒ $CE =\frac{ 87}{ \sqrt3} = 29\sqrt3 \,m$

In ∆BCG,

$\frac{BG}{ CG}= tan 30^{\degree}$

$\frac{ 88.2 - 1.2}{ CG} = \frac{1}{ \sqrt3}$

$87 \sqrt3 m = \frac1{ CG}$

Distance travelled by balloon = EG = CG − CE
= $( 87 \sqrt3 - 29 \sqrt3)\,m$
= $58 \sqrt3 \,m$

Therefore, The distance travelled by balloon is $58 \sqrt3 \,m$.