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, show that the height of the tower is $\frac{d}{\sqrt{co{t}^2\alpha +co{t}^2\beta }}$.

Answer 1

Let PQ be the tower of height h
and AQ = y, BQ = x
Given: $AB=d,\angle PAQ=\alpha \angle PBQ=\beta$
Now in $\Delta AQB$
$A{B}^2=A{Q}^2+B{Q}^2$ (Pythagoras Theorem)
$\Rightarrow d^2=x^2+y^2$ ....(1)
In $\Delta PAQ$
$cot\alpha =\frac{AQ}{PQ}=\frac{y}{h}$
$\Rightarrow y=hcot\alpha$ ....(2)
In $\Delta PBQ$
$cot\beta =\frac{BQ}{PQ}=\frac{x}{h}$
$\Rightarrow x=hcot\beta$ .... (3)
Squaring and adding (2) and (3) we get
$x^2+y^2=h^{2}co{t}^2\alpha +h^{2}cot\beta$
$\Rightarrow d^2=h^2(co{t}^2\alpha +co{t}^2\beta )$ (from (1))
$\Rightarrow h=\sqrt{\frac{d^2}{co{t}^2\alpha +co{t}^2\beta }}=\frac{d}{\sqrt{co{t}^2\alpha +co{t}^2\beta }}$