Question

The angle of elevation of a cloud from a point h metres above a lake is α and the angle of depression of its reflection is β. Prove that the distance of the cloud from the point of observation is $\frac{2h\sec \mathrm{\alpha }}{(\tan \mathrm{\beta }-\tan \mathrm{\alpha })}.$

Answer 1

Let $AB$ be the surface of the lake and $P$ be the point vertically above $A$ such that $AP=h$ m.
Let $C$ be the position of the cloud and $D$ be its reflection on the lake.
Let the height of the cloud be $H$ metres. Thus, we have:
$BC=BD=H$ m
Draw $PQ$⊥$CD$.
Thus, we have:
∠$CPQ=\alpha$ and ∠$QPD=\beta$
$AP=BQ=h$ m
$CQ=(BC-BQ)=(H-h)$ m and $DQ=(H+h)$ m


From the right ∆$CQP$, we have:
$\frac{CQ}{PQ}=\tan \alpha$
⇒ $\frac{(H-h)}{PQ}=\tan \alpha$
⇒ $PQ=\frac{(H-h)}{\tan \alpha }$ ...(i)
From the right ∆$QPD$, we have:
$\frac{DQ}{PQ}=\tan \beta$
⇒ $\frac{(H+h)}{PQ}=\tan \beta$
⇒ $PQ=\frac{(H+h)}{\tan \beta }$ ...(ii)

From (i) and (ii), we get:
$\frac{(H-h)}{\tan \alpha }=\frac{(H+h)}{\tan \beta }$
⇒ $H\tan \beta -h\tan \alpha =H\tan \alpha +h\tan \beta$
⇒ $H(\tan \beta -\tan \alpha )=h(\tan \alpha +\tan \beta )$
⇒ $H=\frac{h(\tan \alpha +\tan \beta )}{(\tan \beta -\tan \alpha )}$ ...(iii)
Using the value of $H$ in (i), we get:
$PQ=\frac{(H-h)}{\tan \alpha }=\frac{H}{\tan \alpha }-\frac{h}{\tan \alpha }=\frac{h(\tan \alpha +\tan \beta )}{\tan \alpha (\tan \beta -\tan \alpha )}-\frac{h}{\tan \alpha }$

$$\begin{gathered} =\frac{h\tan \alpha +h\tan \beta -h\tan \beta +h\tan \alpha }{\tan \alpha (\tan \beta -\tan \alpha )} \\ =\frac{2h\tan \alpha }{\tan \alpha (\tan \beta -\tan \alpha )} \\ =\frac{2h}{(\tan \beta -\tan \alpha )} \end{gathered}$$ ...(iv)

Now, to find $PC$, we have:
$\cos \alpha =\frac{PQ}{PC}$
⇒ $PC=\frac{PQ}{\cos \alpha }=PQsec\alpha$
Putting the value of $PQ$ from (iv), we get:
$PC=\frac{2hsec\alpha }{(\tan \beta -\tan \alpha )}$

Distance of the cloud = $PC=\frac{2hsec\alpha }{(\tan \beta -\tan \alpha )}$