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 ${60}^0$. After some time the angle of elevation reduces to ${30}^0$ (see fig.). Find the distance traveled by the balloon during the interval.

Answer 1

Let eye of $1.2m$ tall girl is $A$. The balloon flying $88.2m$ height from ground level have angle of elevation at different distances are ${60}^0$ and ${30}^0$ respectively.

$\therefore \angle BAE={60}^0$ and $\angle DAC={30}^0$

$BE=CD=88.2–1.2=87m$

From right angled $\Delta ABE$

$\tan {60}^0=\frac{BE}{AB}$

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

$x=\frac{87}{\sqrt{3}}$..................(i)

Fro right angled $\Delta ACD$

$\tan {30}^0=\frac{CD}{AC}$

$\frac{1}{\sqrt{3}}=\frac{87}{x+y}$

$x+y=87\sqrt{3}$

$x=87\sqrt{3}-y$...........................(ii)

From equation (i) and (ii)

$\frac{87}{\sqrt{3}}=87\sqrt{3}-y$

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

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

$\sqrt{3}y=261-87$

$\sqrt{3}y=174$

$y=\frac{174}{\sqrt{3}}$

$y=\frac{174}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$

$y=\frac{174\sqrt{3}}{3}$

$y=58\sqrt{3}m$

Hence, distance traveled by balloon = $58\sqrt{3}m$