Question

The angles of elevation of the top of a tower from two points A and B at a distance of a and b respectively from the base and in the same straight line with it are complementary. Prove that the height of the tower is `sqrt(ab)`.

Answer 1

Let the height of the tower 'OT' = h.
Let O be the base of the tower.
Let A and B be two points on the same line through the base such that
OA = a, OB = b.

∵ The angles at A and B are complementary.

∴ ∠TAO = α
then ∠TBO = 90° - α

In right-angled ΔOAT, 

tan α = `"OT"/"OA" = h/a`         .....(i)

In right-angled ΔOBT, 

tan(90° - α) = `"OT"/"OB" = h/b`       

cot α = h/b`   .....(ii)

Multiplying (i) and (ii), we have

tan α cot α = `h/a xx h/b = h^2/(ab)`

⇒ 1 = `h^2/(ab)`

⇒ h² = ab

⇒ h = `sqrtab`

Hence, the height of the tower = `sqrtab`.