The angle of elevation of a stationary cloud from a point 25 m above a lake is 30° and the angle of depression of its reflection in the lake is 60°. What is the height of the cloud above the lake-level?

Let C be the position of the cloud, l be the surface of the lake and D be reflection of the cloud.

Let AB be x and CB be h,

In △ABC,

$\Rightarrow \tan 30° = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{BC}{AB} \\[1em] \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{h}{x} \\[1em] \Rightarrow \sqrt3h = x ….(1)$

In △ABD,

$\Rightarrow \tan 60° = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{BD}{AB} \\[1em] \Rightarrow \sqrt3 = \dfrac{25 + 25 + h}{x} \\[1em] \Rightarrow \sqrt3x = 50 + h ….(2)$

Substituting value of x from equation (1) in (2), we get :

$\Rightarrow \sqrt3(\sqrt3h) = 50 + h$

⇒ 3h = 50 + h

⇒ 2h = 50

⇒ h = 25 m.

Height of cloud above lake-level = OC = 25 + h = 25 + 25 = 50 m.

Hence, height of the cloud above lake-level is 50 m.