Question

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

Answer 1

Let a be a point h metres above the lake AF and B be the position of the cloud.
Draw a line parallel to EF from A on BD at C.

But, BF = DF

Let, BC = m

so, BF = (m + h)

⇒ BF = DF = (m + h) metres

Consider ΔBAC,

$AB=mcosec\alpha$ ----(1)
and
$AC=mcot\alpha$

Consider $△ACD$

$AC=(2h+m)cot\beta$

Therefore $(2h+m)cot\beta =mcot\alpha$

$m=\frac{2hcot\beta }{(cot\alpha -cot\beta )}$

Substitute this value of m in (1) we get,

$AB=cosec\alpha \times \left[\frac{2hcot\beta }{(cot\alpha -cot\beta )}\right]=\frac{2hsec\alpha }{(tan\beta -tan\alpha )}$

Hence proved.