Question

If the angle of elevation of a cloud from a point h meter above a lake has measure $\alpha$ and the angle depression of its reflection of in the lake has measure$\beta$ . Prove that the height of the cloud is $\frac{h(tan\beta +tan\alpha )}{tan\beta -tan\alpha }$

Answer 1

Let AB be the surface of the lake and let P be a point of observation such that AP = h metres.

Let C be the position of the cloud and C' be its reflection in the lake.

Then, CB=C'B. Let PM be perpendicular from P on CB. Then, $\angle$CPM= $\alpha$ and $\angle$MPC' = $\beta$. Let CM = x.

Then, CB=CM+MB=CM+PA = x+h

In $△$ CPM, we have,

$tan\alpha =\frac{CM}{PM}$

$tan\alpha =\frac{x}{AB}$

$AB=xcot\alpha$ .....(1)

In $△$PMC', we have

$tan\beta =\frac{C^{\prime }M}{PM}$

$tan\beta =\frac{x+2h}{AB}$

$AB=(x+2h)cot\beta$ .....(2)

From (1) & (2), we have,

$xcot\alpha =(x+2h)cot\beta$

$x(cot\alpha -cot\beta )=2hcot\beta$

$x(\frac{1}{tan\alpha }-\frac{1}{tan\beta })=\frac{2h}{tan\beta }$

$x\left(\frac{tan\beta -tan\alpha }{tan\alpha tan\beta }\right)=\frac{2h}{tan\beta }$

$x=\frac{2htan\alpha }{tan\beta -tan\alpha }$

Hence, the height CB of the cloud is given by

$CB=x+h$

$CB=\frac{2htan\alpha }{tan\beta -tan\alpha }+h$

$CB=\frac{2htan\alpha +htan\beta -htan\alpha }{tan\beta -tan\alpha }=\frac{h(tan\alpha +tan\beta )}{tan\beta -tan\alpha }$