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 tower is `beta`. If `AB=d`, show that the height of the tower is `d/sqrt(cot^2 alpha+ cot^2 beta)`.

Explanation

Let OP be the tower and let A and B be two points due south and east
respectively of the tower such that
`∠OAP=α`

and `∠OBP=β`.

Let OP=h

In `ΔOAP`,we have `tanα=h/(OA)` or ​`OA=hcotα.`.............(i)

In `ΔOBP` ,we have `tanβ=h/(OB)`or `OB=hcotβ`..............(ii)
​
Since OAB is a right angled triangle .

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