if the angle of elevation of a cloud from a point ' h ' metres above a lake is alpha and the angle of depression of its reflection in the lake is beeta. prove that the distance of the cloud from the point of observation is 2h sec alpha
tan beeta - tan alpha??
Let a be a point h metres above the pond AF and B be the position of the Sky.
We draw a line parallel to EF from A on BD at C.
Clearly, BF = DF
Let, BC = m
so, BF = (m + h)
=> BF = DF = (m + h) metres
Now, in triangle BAC we have,
AB = m cosec α ---- (i)
and, AC = m cot α
Again,
in triangle ACD we have,
AC = (2h + m) cot β
so, m cot α = (2h + m) cot β
=> m = 2h cot β / (cot α - cot β)
putting, the value of m in (i) we get,
AB = cosec α [2h cot β / (cot α - cot β)] = 2h sec α / (tan β - tan α) proved..!!