From the top of a hill h metres high,the angl | vedclass

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(A) Let $AB$ be a hill of height $h$ metres and $CD$ be a pillar of height $h'$ metres. Let $E$ be a point on $AB$ such that $ED$ is horizontal.In $

Vedclass · Accepted Answer

$\frac{h(	an \beta-	an \alpha)}{	an \beta}$

Vedclass · Answer

(A) Let $AB$ be a hill of height $h$ metres and $CD$ be a pillar of height $h'$ metres. Let $E$ be a point on $AB$ such that $ED$ is horizontal.In $	riangle BED$,$	an \alpha = \frac{BE}{ED} = \frac{h-h'}{ED} \implies ED = \frac{h-h'}{	an \alpha}$.In $	riangle BAC$,$	an \beta = \frac{AB}{AC} = \frac{h}{ED} \implies ED = \frac{h}{	an \beta}$.Equating the two expressions for $ED$:$\frac{h-h'}{	an \alpha} = \frac{h}{	an \beta}$$h-h' = \frac{h 	an \alpha}{	an \beta}$$h' = h - \frac{h 	an \alpha}{	an \beta} = h \left(1 - \frac{	an \alpha}{	an \beta}ight) = \frac{h(	an \beta - 	an \alpha)}{	an \beta}$.

From the top of a hill $h$ metres high,the angles of depression of the top and the bottom of a pillar are $\alpha$ and $\beta$ respectively. The height (in metres) of the pillar is

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