Question

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

Answer 1

Let AB be the surface of the lake and let C be a point of observation such that AC = h metres Let D be the position of the cloud and D be its reflection in the lake Then BD = BD
In $\Delta$ DCE
$\tan \alpha =\frac{DE}{CE}\Rightarrow CE=\frac{H}{\tan \alpha }............\left(i\right)$
In $\Delta$ CED'
$\Rightarrow CE=\frac{h+H+h}{\tan \beta }\Rightarrow CE=\frac{2h+H}{\tan \beta }............\left(ii\right)$
From (i) & (ii)
$\Rightarrow \frac{H}{\tan \alpha }=\frac{2h+H}{\tan \beta }\Rightarrow H\tan \beta =2h\tan \alpha +H\tan \alpha$
$H\tan \beta -H\tan \alpha =2h\tan \alpha \Rightarrow H(\tan \beta -\tan \alpha )=2h\tan \alpha$
$\Rightarrow H=\frac{2h\tan \alpha }{\tan \beta -\tan \alpha }.........\left(iii\right)$
In $\Delta$ DCE $\sin \alpha =\frac{DE}{CD}\Rightarrow CD=\frac{DE}{\sin \alpha }\Rightarrow \Rightarrow CD=\frac{H}{\sin \alpha }$
Substituting the value of H from (iii)
$CD=\frac{2h\tan \alpha }{(\tan \beta -\tan \alpha )\sin \alpha }\Rightarrow CD=\frac{2h\frac{\sin \alpha }{\cos \alpha }}{(\tan \beta -\tan \alpha )\sin \alpha }$
$CD=\frac{2h\sec \alpha }{\tan \beta -\tan \alpha }$
Hence the distance of the cloud from the point of observation is $\frac{2h\sec \alpha }{\tan \beta -\tan \alpha }$