If the angle of elevation of a cloud from a point 100 m above a lake is 45° and the angle of depression of its reflection in the lake is 60°. Then the height of the cloud above the lake is?

50(√3 + 3) m

Here, BD = 100m, let, AD = (x) m,  DC = 100 + x

In Δ PDC;

tan 60° = \(\frac{DC}{PD}\)

⇒ \(\frac{\sqrt {3}}{1}\) = \(\frac{100 + x}{PD}\)  .......(i)

and

In Δ PDA;

tan 45° = \(\frac{AD}{PD}\)

⇒ \(\frac{1}{1}\) = \(\frac{x}{PD}\)   .........(ii)

Dividing (i) by (ii), we get,

⇒ \(\sqrt {3}\) = \(\frac{100 + x}{x}\) 

⇒ x (\(\sqrt {3}\) - 1) = 100

⇒ x = \(\frac{100}{\sqrt {3} - 1}\) = 50 (\(\sqrt {3}\) + 1)

Height of cloud (AB) = AD + DB = (x + 100) = 50 (\(\sqrt {3}\) + 1) + 100

                                        =  50 (\(\sqrt {3}\) + 3) m