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°. Find the distance travelled by the balloon during the interval.

From figure,

FE = FD - ED = FD - AB = 88.2 - 1.2 = 87 m.

GH = FE = 87 m

In △AGH,

tan 60° = $\dfrac{\text{Side opposite to angle 60°}}{\text{Side adjacent to angle 60°}}$

Substituting values we get :

$\Rightarrow \sqrt{3} = \dfrac{GH}{AH} \\[1em] \Rightarrow AH = \dfrac{GH}{\sqrt{3}} \\[1em] \Rightarrow AH = \dfrac{87}{\sqrt{3}} \text{ m}.$

In △AEF,

tan 30° = $\dfrac{\text{Side opposite to angle 30°}}{\text{Side adjacent to angle 30°}}$

Substituting values we get :

$\Rightarrow \dfrac{1}{\sqrt{3}} = \dfrac{FE}{AE} \\[1em] \Rightarrow AE = FE\sqrt{3} = 87\sqrt{3} \text{ m}.$

From figure,

EH = AE - AH

= $87\sqrt{3} - \dfrac{87}{\sqrt{3}}$

= $\dfrac{87 \times 3 - 87}{\sqrt{3}}$

= $\dfrac{261 - 87}{\sqrt{3}} = \dfrac{174}{\sqrt{3}}$

= $58\sqrt{3}$ m.

Hence, distance travelled by the balloon during the interval = $58\sqrt{3}$ m.