The angle of elevation of a cloud at $a$ height $h$ above the level of water in a lake is $α$
and the angle of depression of its image in the lake is $β$ . The height of the
cloud above the surface of the lake is

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$\frac{h (\tan α - \tan β)}{\tan α + \tan β}$

$\frac{h (\cot α + \cot β)}{\cot β - \cot α}$

$\frac{h \sin (α + β)}{\sin (β - α)}$

$\frac{h \sin (α - β)}{\sin (α + β)}$

answer is C.

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Detailed Solution

Let the height of the cloud above the water
level be $x$ . Let $A$ be the point at $a$ height $h$ above the water
level from which the cloud’s angles of depression and
elevation are measured. Let $L$ and $L'$ be the locations of
the cloud and its image, respectively. Further, as shown in
Fig. 27.31, let $P$ be the point directly below the cloud, midway
between $L$ and $L'$ and $Q$ the point directly above $P$ at $a$
height $h$ . Then $∠ L A Q = α, ∠ Q A L' = β, P Q = h, Q L = x - h$ ,
$P L' = x$ , and we have $(x - h) c o t a = (x + h) c o t b = A Q$ ,

i.e.,

$\frac{x + h}{x - h} = \frac{\tan β}{\tan α} \Rightarrow \frac{2 x}{2 h} = \frac{\tan β + \tan α}{\tan β - \tan α}$

$\begin{array}{l} \Rightarrow x = h (\frac{\tan β + \tan α}{\tan β - \tan α}) = h (\frac{\cot α + \cot β}{\cot α - \cot β}) \\ = h [\frac{\sin (α + β)}{\sin (β - α)}] \end{array}$