Question

If the angle of elevation of a cloud from a point $h$ metres above a lake be $\theta$ and the angle of depression of its reflection in the lake be $\varphi$, prove that the distance of the cloud from the point of observation is $\frac{2h\cos \varphi }{\sin (\varphi -\theta )}$
Also find the horizontal distance of the cloud from the place of observation.

Answer 1

we have to find $PL$. Replacing $\alpha$ by $\theta$ and $\beta$ by $\varphi$, we have
$x+h=y\tan \varphi$
$x-h=y\tan \theta$
Subtract $2h=y(\tan \varphi -\tan \theta )=\frac{y\sin (\varphi -\theta )}{\cos \varphi \cos \theta }$
$\therefore y=\frac{2h\cos \theta \cos \varphi }{\sin (\varphi -\theta )}$
Above gives, the horizontal distance of the cloud from the point of observation.
Again $y=PL=\cos \theta$
$\therefore PL=\frac{2h\cos \varphi }{\sin (\varphi -\theta )}$