Question

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

Answer 1

Let $OP$ be the tower and let $A$ and $B$ be two points due south and east

respectively of the tower such that

$\angle OAP=\alpha$

and $\angle OBP=\beta$.

Let $OP=h$

$\begin{array}{c}In\Delta OAP,wehave\\ \tan \alpha =\frac{h}{OA}\\ OA=h\cot \alpha ..............\left(i\right)\\ In\Delta OBP,wehave\\ \tan \beta =\frac{h}{OB}\\ OB=h\cot \beta ..............\left(i\right)\end{array}$

Since $OAB$ is a right angled triangle .

Therefore ,

$\begin{array}{c}A{B}^2=O{A}^2+O{B}^2\\ d^2=h^2+{\cot }^2\alpha +h^2+{\cot }^2\beta \\ \therefore h=\frac{d}{\sqrt{{\cot }^2\alpha +{\cot }^2\beta }}\left[u\sin g\left(i\right)and\left(ii\right)\right]\end{array}$