Question

The angle of elevation of a stationary cloud from a point $2500cm$ above a lake is ${15}^o$ and the angle of depression of its reflection in the lake is ${45}^o$. What is the height of the cloud above the lake level ?

Answer 1

Proceed as above, $h=2500,\beta ={45}^o,\alpha ={15}^o$.
$\therefore x=2500\times \frac{\sin (45-15)}{\sin (45-15)}=2500\times \frac{\sin 60}{\sin 30}=2500\left(\sqrt{3}\right)$ metres.