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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)`.
Solution
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`.
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