Q67 of 107 Page 12

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.

In the fig let AB is the Tower of height h (m)


In ∆ABD


tan α =


tan α =


h = 4 tan α …………(1)


In ∆ABC


tan (90°-α)=


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


h = 6m


Therefore height of the tower is 6m



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