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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.
Solution
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.
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