From a point a metre above a lake,the angle o | vedclass

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(B) Let the point of observation be $P$ at a height $a$ above the lake surface. Let the cloud be at $C$ at a height $H$ above the lake surface. The re

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$\frac{a \sin(\alpha + \beta)}{\sin(\beta - \alpha)} 	ext{ metre}$

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(B) Let the point of observation be $P$ at a height $a$ above the lake surface. Let the cloud be at $C$ at a height $H$ above the lake surface. The reflection of the cloud is at $C'$ at a depth $H$ below the lake surface.In the right-angled triangle formed by the cloud,the point $P$,and the horizontal line,we have $	an \alpha = \frac{H - a}{x}$,where $x$ is the horizontal distance.So,$x = (H - a) \cot \alpha$.In the right-angled triangle formed by the reflection $C'$,the point $P$,and the horizontal line,we have $	an \beta = \frac{H + a}{x}$.So,$x = (H + a) \cot \beta$.Equating the two expressions for $x$: $(H - a) \cot \alpha = (H + a) \cot \beta$.$H \cot \alpha - a \cot \alpha = H \cot \beta + a \cot \beta$.$H(\cot \alpha - \cot \beta) = a(\cot \alpha + \cot \beta)$.$H \left( \frac{\cos \alpha}{\sin \alpha} - \frac{\cos \beta}{\sin \beta} ight) = a \left( \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \beta}{\sin \beta} ight)$.$H \left( \frac{\sin \beta \cos \alpha - \cos \beta \sin \alpha}{\sin \alpha \sin \beta} ight) = a \left( \frac{\sin \beta \cos \alpha + \cos \beta \sin \alpha}{\sin \alpha \sin \beta} ight)$.$H \sin(\beta - \alpha) = a \sin(\alpha + \beta)$.$H = \frac{a \sin(\alpha + \beta)}{\sin(\beta - \alpha)}$.

From a point $a$ metre above a lake,the angle of elevation of a cloud is $\alpha$ and the angle of depression of its reflection is $\beta$. The height of the cloud is

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