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The angle of elevation of the top of a t...

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)`.

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`tan(alpha)=h/(d1) & tan(beta)=h/(d2)` then, `d1=hcot(alpha)` `d2=hcot(beta)`
Since, `/_AOB` is right angle triangle at o. So, `d1^2 + d2^2 = d^2` `(hcot(alpha))^2 + (hcot(beta))^2= d^2` Therefore, `h= d/sqrt((cot(alpha)^2)+(cot(beta))^2)`
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