Question

The angle of elevation of an aeroplane from a point $P$ on the ground is ${60}^{\circ }$. After $12$ seconds, the angle of elevation changes to ${30}^{\circ }$. If the plane is flying horizontally at a speed of $600\sqrt{3}km/h$, find the height at which its is flying.

Answer 1

Let $A$ be the position of the plane, and it moves towards $C$. Let $BP=x$. Time taken by plane to reach $C$ is $12s=\frac{12}{60\times 60}hr$

$Distance=Speed\times Time$

$AC=600\sqrt{3}\times \frac{12}{60\times 60}=2\sqrt{3}km$

In $△ABP,\frac{AB}{BP}=\tan {60}^{\circ }⟹\frac{AB}{x}=\sqrt{3}$

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

From figure, $BD=AC=2\sqrt{3}km$

$PD=PB+BD=\frac{AB}{\sqrt{3}}+2\sqrt{3}=\frac{AB+6}{\sqrt{3}}km$

In $△CDP,\frac{CD}{PD}=\tan {30}^{\circ }$

$\frac{AB}{\frac{AB+6}{\sqrt{3}}}=\frac{1}{\sqrt{3}}$

$3AB=AB+6$

$AB=3km$. This is the height of the plane.