Question

In the angle of elevation of a cloud from a point $h$ metres above a lake be $\beta$, and the angle of depression of its reflection in the lake be $\alpha$, prove that the height of the cloud is $h\left(\frac{\tan \alpha +\tan \beta }{\tan \alpha -\tan \beta }\right)$.

Answer 1

$\begin{array}{c}\tan \alpha =\frac{x}{OM}\\ \Rightarrow OM=\frac{x}{\tan \alpha }\to \left(i\right)\\ =\tan \beta =\frac{2h+x}{OM}\\ OM=\frac{2h+x}{\tan \beta }\to \left(ii\right)\\ fromequation\left(i\right)and\left(ii\right)\\ \Rightarrow \frac{x}{\tan \alpha }=\frac{2h+x}{\tan \beta }\\ \Rightarrow x\tan \beta =\left(2h+x\right)\tan \alpha \\ \Rightarrow x=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }\\ heightofcloud=h+x\\ =h+\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }=\frac{h\left(\tan \alpha +\tan \beta \right)}{\tan \beta -\tan \alpha }\end{array}$