(C) Let $h$ be the height of the tower $OP$,where $O$ is the base of the tower.In $\triangle OAP$,$\tan \alpha = \frac{h}{OA} \Rightarrow OA = h \cot

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$\frac{d}{\sqrt{\cot^2 \alpha + \cot^2 \beta}}$

(C) Let $h$ be the height of the tower $OP$,where $O$ is the base of the tower.In $\triangle OAP$,$\tan \alpha = \frac{h}{OA} \Rightarrow OA = h \cot \alpha$.In $\triangle OBP$,$\tan \beta = \frac{h}{OB} \Rightarrow OB = h \cot \beta$.Since $OA$ is south and $OB$ is east,$\triangle OAB$ is a right-angled triangle at $O$.By Pythagoras theorem in $\triangle OAB$,$OA^2 + OB^2 = AB^2$.$(h \cot \alpha)^2 + (h \cot \beta)^2 = d^2$.$h^2 (\cot^2 \alpha + \cot^2 \beta) = d^2$.$h^2 = \frac{d^2}{\cot^2 \alpha + \cot^2 \beta}$.$h = \frac{d}{\sqrt{\cot^2 \alpha + \cot^2 \beta}}$.

Class 11 Mathematics Trigonometrical Equations Height and Distance

Medium

The angle of elevation of the top of a tower from a point $A$ due south of the tower is $\alpha$ and from a point $B$ due east of the tower is $\beta$. If $AB = d$,then the height of the tower is

A
$\frac{d}{\sqrt{\tan^2 \alpha - \tan^2 \beta}}$
B
$\frac{d}{\sqrt{\tan^2 \alpha + \tan^2 \beta}}$
C
$\frac{d}{\sqrt{\cot^2 \alpha + \cot^2 \beta}}$
D
$\frac{d}{\sqrt{\cot^2 \alpha - \cot^2 \beta}}$

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The angle of elevation of the top of a tower from a point $A$ due south of the tower is $\alpha$ and from a point $B$ due east of the tower is $\beta$. If $AB = d$,then the height of the tower is

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