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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