Question

If the angle of elevation of a cloud from a point 200 m above a lake is ${30}^{\circ }$ and the angle of depression of its reflection in the lake is ${60}^{\circ }$, then the height of the cloud above the lake is ___.

Answer 1

Let AB = H m be the height of the cloud above the lake. BC is the lake surface and BQ is the distance of the reflection formed from the lake surface.


AB = BQ = H m, CD = PB = 200 m
AP = AB - PB = H - 200
PQ = PB + BQ = 200 + H

in $\Delta ADP,tan{30}^{\circ }=\frac{AP}{PD}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{H-200}{PD}\Rightarrow PD=\sqrt{3}(H-200)$

In $\Delta PDQ,tan{60}^{\circ }=\frac{PQ}{PD}$
$\Rightarrow \sqrt{3}=\frac{H+200}{PD}\Rightarrow PD=\frac{H+200}{\sqrt{3}}$


$(H-200)\sqrt{3}=\frac{H+200}{\sqrt{3}}$
$\Rightarrow$ 3(H - 200) = H + 200
$\Rightarrow$ 3H - 600 = H + 200
$\Rightarrow$ 2H = 800 $\Rightarrow$ H = 400 m