The angle of elevation of a cloud from a point $h$ metres above the surface of a lake is $\theta$ and the angle of depression of its reflection in the lake is $\phi$. Prove that the height of the cloud above the lake is $h\left(\frac{\tan \phi+\tan \theta}{\tan \phi-\tan \theta}\right).$

(N/A) Let $P$ be the cloud and $Q$ be its reflection in the lake. Let $A$ be the point of observation such that $AB = h$. Let the height of the cloud above the lake be $x$. Let $AL = d$.
From $\triangle PAL$,we have $\frac{x-h}{d} = \tan \theta$ ..........$(1)$
From $\triangle QAL$,we have $\frac{x+h}{d} = \tan \phi$ ..........$(2)$
Dividing $(2)$ by $(1)$,we get $\frac{x+h}{x-h} = \frac{\tan \phi}{\tan \theta}$.
Applying componendo and dividendo,we get $\frac{(x+h)+(x-h)}{(x+h)-(x-h)} = \frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}$.
This simplifies to $\frac{2x}{2h} = \frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}$.
Therefore,$x = h\left(\frac{\tan \phi + \tan \theta}{\tan \phi - \tan \theta}\right).$