Question

A tower subtends an angle $\alpha$ at a point A in the plane of its base and the angle of depression of the foot of the tower at a point b meters just above A is $\beta$. Prove that the height of the tower is b tan $\alpha$ cot $\beta$.

Answer 1

Let x be the distance of the point A from the foot of the tower and h be the height
of the tower.
In △APQ,
h = x tanα ...(i)
In △PRB
b = x tanβ ...(ii)
From (i) and (ii),
h = $=\frac{btan\alpha }{tan\beta }$ = b.tanα cotβ
Therefore the height of the tower is b tanα cotβ.