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The top of a hill observed from the top and bottom of a building of height $\mathrm{h}$ is at angles of elevation $\alpha$ and $\beta$ respectively. The height of the hill is: $\quad$
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The correct answer is:
$\frac{\mathrm{h} \cot \alpha}{\cot \alpha-\cot \beta}$
$\operatorname{In} \Delta \mathrm{BDC}$,
$\tan \beta=\frac{\mathrm{CD}}{\mathrm{BD}}=\frac{\mathrm{H}}{\mathrm{BD}}$
$\mathrm{BD}=\frac{\mathrm{H}}{\tan \beta}=\mathrm{H} \cot \beta$
In $\triangle$ ALC, $\tan \alpha=\frac{\mathrm{CL}}{\mathrm{AL}}=\frac{\mathrm{H}-\mathrm{h}}{\mathrm{BD}}$
$\mathrm{BD}=(\mathrm{H}-\mathrm{h}) \cot \alpha$
from (i) \& (ii), $(\mathrm{H}-\mathrm{h}) \cot \alpha=\mathrm{H} \cot \beta$
$\mathrm{H} \cot \alpha-\mathrm{h} \cot \alpha=\mathrm{H} \cot \beta$
$\mathrm{H}[\cot \alpha-\cot \beta]=\mathrm{h} \cot \alpha$
$\mathrm{H}=\frac{\mathrm{h} \cot \alpha}{\cot \alpha-\cot \beta}$
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