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 ${60}^{\circ }$. After some time, the angle of elevation reduces to ${30}^{\circ }$. Find the distance travelled by the balloon during the interval.

Answer 1

Let the initial position of the balloon be A and final position be B.
Height of balloon above the girl height $=88.2m-1.2m=87m$
Distance travelled by the balloon = $DE=CE-CD$
As per question,
In right ΔBEC,
tan ${30}^{\circ }=\frac{BE}{CE}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{87}{CE}$
$\Rightarrow CE=87\sqrt{3}m$
Also,
In right $\Delta ADC,$
tan ${60}^{\circ }=\frac{AD}{CD}$
$\Rightarrow \sqrt{3}=\frac{87}{CD}$
$\Rightarrow CD=\frac{87}{\sqrt{3}}m=29\sqrt{3}m$
Distance travelled by the balloon$=DE=CE-CD=(87\sqrt{3}-29\sqrt{3})m=29\sqrt{3}(3-1)m$ =100.45 m