Question

If the angle of elevation of a cloud from a point $h$ meters above a lake is $\alpha$ and the angle of depression of its reflection in the lake be $\beta$, prove that the distance of the cloud from the point of observation is $\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }$.

Answer 1

Let $OB=x$

Now $\tan \alpha =\frac{\gamma }{x}$ ________ (1)

$\tan \beta =\frac{h+h+y}{x}$

$\therefore \tan \beta =\frac{2h+y}{x}$ _______ (2)

$\therefore \frac{\tan \alpha }{\tan \beta }=\frac{\gamma }{2h+y}$

$\therefore 2h\tan \alpha +y\tan \alpha =\gamma \tan \beta$

$\therefore 2h\tan \alpha =\gamma (\tan \beta -\tan \alpha )$

$\therefore \gamma =\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }$

Hence proved.