Question

The angles of elevation of the top of a tower from two points on the ground at distances a metres and b metres from the base of the tower and in the sqme straight line with it are complementary. Prove that the he8ght of the tower is √ab metres

Answer 1

Here is your answer

Answer 2

From the Figure,

In ΔACD,

TanΘ = CD/AD

          = CD/a                          ................(i)
 
In ΔBCD,

Tan(90-Θ) = CD/BD
 
                  = CD/b                  ................(ii)

Multiplying (i) and (ii)

TanΘ × Tan(90-Θ) = CD/a × CD/b

TanΘ × CotΘ = CD²/ab

(because Tan(90-Θ) = CotΘ)

1 = CD²/ab

(because TanΘ × CotΘ = 1)

CD² = ab

CD = √ab metres

Hence the Height = √ab metres

Hence Proved.
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