Question

# A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b feet just above A is $\beta$. Then, the height of the tower is

A

$b\mathrm{tan}\left(\alpha \right)cot\left(\beta \right)$

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B

$bcot\left(\alpha \right)\mathrm{tan}\left(\beta \right)$

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C

$bcot\left(\alpha \right)cot\left(\beta \right)$

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D

$b{\mathrm{tan}}^{2}\left(\alpha \right)cot\left(\beta \right)$

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Solution

## The correct option is A $b\mathrm{tan}\left(\alpha \right)cot\left(\beta \right)$Finding the height of the tower:Given, A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b feet just above A is $\beta$. Assume ‘X’ is the distance from point ‘A’ to foot of the tower.Let ‘h’ be the height of the tower $\mathrm{h}=\mathrm{PQ}$In $△\mathrm{APQ},$$⇒\mathrm{h}=\mathrm{xtan}\left(\mathrm{\alpha }\right).....\left(\mathrm{i}\right)$In $△\mathrm{PRB}$$\begin{array}{rcl}& ⇒& \mathrm{b}=\mathrm{xtan}\left(\mathrm{\beta }\right)\\ & ⇒& \mathrm{x}=\frac{\mathrm{b}}{\mathrm{tan}\left(\mathrm{\beta }\right)}\\ & ⇒& x=bcot\left(\beta \right)...\left(\mathrm{ii}\right)\end{array}$From equation $\left(i\right)&\left(ii\right)$$⇒\mathrm{h}=\mathrm{btan}\left(\mathrm{\alpha }\right)\mathrm{cot}\left(\mathrm{\beta }\right)$Hence, correct option is $\mathbf{\left(}\mathbit{A}\mathbf{\right)}\mathbf{.}$

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