Q. 674.1( 31 Votes )

The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer :


In the fig let AB is the Tower of height h (m).
Since the tower is vertical to the ground.

 ∠ABC = 90°
In a right-angled triangle, we know,

In ∆ABD


tan α =


tan α =


h = 4 tan α …………(1)


In ∆ABC


tan (90°-α)=

We know tan(90°- θ) = cot θ

⇒ cot α =


⇒  h = 9 cot β ……….(2)


On multiplying eqn (1) and eqn (2), we get


h × h = 4 tan α × 9 cot α

h2 = 36 tan α ×  cot α
We know,


h2 = 36
h = ± 36

As the height cannot be negative.

⇒ h = 6m

Therefore the height of the tower is 6m.

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