if the angle of elevation of a cloud from a point h metres above a lake is a and the angles of depression of reflection is the lack beta prove that the distance of cloud from the point of the observation is (2h sec a) /( tanb-tana
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Let a be a point h metres above the lake AF and B be the position of the cloud.
Draw a line parallel to EF from A on BD at C.
But, BF = DF
Let, BC = m
so, BF = (m + h)
⇒ BF = DF = (m + h) metres
Consider ΔBAC,
AB = m cosec α ---------- (1)
and, AC = m cot α
Consider ΔACD,
AC = (2h + m) cot β
Therefore, m cot α = (2h + m) cot β
⇒ m = 2h cot β / (cot α - cot β)
Substituting the value of m in (1) we get,
AB = cosec α [2h cot β / (cot α - cot β)] = 2h sec α / (tan β - tan α)
Hence proved.
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