(A) Let $AB$ be a hill of height $h$ and $CD$ be a pillar of height $h^{\prime}$.Let $E$ be a point on $AB$ such that $ED$ is horizontal.In $\triangle

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

(A) Let $AB$ be a hill of height $h$ and $CD$ be a pillar of height $h^{\prime}$.Let $E$ be a point on $AB$ such that $ED$ is horizontal.In $\triangle EBD$,$\tan \alpha = \frac{BE}{ED} = \frac{h-h^{\prime}}{ED}$,so $ED = \frac{h-h^{\prime}}{\tan \alpha}$.In $\triangle ABC$,$\tan \beta = \frac{AB}{AC} = \frac{h}{ED}$,so $ED = \frac{h}{\tan \beta}$.Equating the two expressions for $ED$:$\frac{h-h^{\prime}}{\tan \alpha} = \frac{h}{\tan \beta}$$h-h^{\prime} = \frac{h \tan \alpha}{\tan \beta}$$h^{\prime} = h - \frac{h \tan \alpha}{\tan \beta} = h \left(1 - \frac{\tan \alpha}{\tan \beta}\right) = \frac{h(\tan \beta - \tan \alpha)}{\tan \beta}$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Medium

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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A
$\frac{h(\tan \beta-\tan \alpha)}{\tan \beta}$
B
$\frac{h(\tan \alpha-\tan \beta)}{\tan \alpha}$
C
$\frac{h(\tan \beta+\tan \alpha)}{\tan \beta}$
D
$\frac{h(\tan \beta+\tan \alpha)}{\tan \alpha}$

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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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