The angle of elevation of a cliff at a point | vedclass

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(C) Let the height of the cliff be $OP = h$. Let $O$ be the base of the cliff on the ground.From the given information,$AB = 100 \ m$ and $AB$ is vert

Vedclass · Accepted Answer

$\frac{100 \cot \beta}{\cot \beta - \cot \alpha}$

Vedclass · Answer

(C) Let the height of the cliff be $OP = h$. Let $O$ be the base of the cliff on the ground.From the given information,$AB = 100 \ m$ and $AB$ is vertical,so $OC = 100 \ m$.Then $CP = OP - OC = h - 100$.In $	riangle AOP$,$	an \alpha = \frac{OP}{OA} = \frac{h}{OA} \implies OA = h \cot \alpha$.In $	riangle BCP$,$	an \beta = \frac{CP}{BC} = \frac{h - 100}{BC}$. Since $BC = OA$,we have $BC = h \cot \alpha$.Substituting $BC$ in the second equation: $	an \beta = \frac{h - 100}{h \cot \alpha}$.$h \cot \alpha 	an \beta = h - 100$.$100 = h - h \cot \alpha 	an \beta = h(1 - \cot \alpha 	an \beta)$.$100 = h \left(1 - \frac{\cot \alpha}{\cot \beta}ight) = h \left(\frac{\cot \beta - \cot \alpha}{\cot \beta}ight)$.Therefore,$h = \frac{100 \cot \beta}{\cot \beta - \cot \alpha}$.

The angle of elevation of a cliff at a point $A$ on the ground and a point $B$,$100 \ m$ vertically above $A$,are $\alpha$ and $\beta$ respectively. The height of the cliff is

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