Question

From a point P on the ground, the angle of elevation of the top of a $10m$ tall building and a helicopter, hovering over the top of the building, are ${30}^{\circ }$ and ${60}^{\circ }$ respectively. Find the height of the helicopter above the ground.

Answer 1

Given $RQ=10m,\angle QPR={30}^o,\angle QPS={60}^o$, let $SR=x$

From $△PQR$, we have

$\tan {30}^{\circ }=\frac{QR}{PQ}$

$PQ=\frac{QR}{\tan {30}^{\circ }}$

$PQ=\frac{10}{\frac{1}{\sqrt{3}}}$

$PQ=10\sqrt{3}..............\left(1\right)$

Similarly,In $△PQS$, we have

$PQ=\frac{SQ}{\tan {60}^{\circ }}$

$PQ=\frac{10+x}{\sqrt{3}}.............\left(2\right)$

From eq.(1) and (2), we get

$10\sqrt{3}=\frac{10+x}{\sqrt{3}}$

$\Rightarrow 10\sqrt{3}\times \sqrt{3}=10+x$

$\Rightarrow 30=10+x$

$\Rightarrow x=20m$

So, Height of helicopter above the ground $=QR+SR=10+20=30m$