Question

The angles of elevation of the top of a tower from two points on the ground at distances a and b metres from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is `sqrt(ab)` metre.

Answer 1

Let AB be the tower of height h metre

Let C and D be two points on the level ground such that BC = b metres, BD = a metres, ∠ACB = α and ∠ADB = β

Given, α + β = 90°

In ΔABC, 

`(AB)/(BC) = tan alpha`

`=> h/b = tan alpha`   ...(i)

In ΔABD, 

`(AB)/(BD) = tan beta`

`=> h/a = tan (90^circ - alpha) = cot alpha`  ...(ii)

Multiplying (i) by (ii), we get, 

`(h/a)(h/b) = 1`

`=>` h² = ab

∴ `h = sqrt(ab)` metre

Hence, height of the tower is `sqrt(ab)` metre.