Angle of elevation = α
Angle of depression of its reflection in lake = β
Height of the cloud above the lake = h m
Proof:
Let AB be the surface of the lake and let P be the point of the observation such that AP = h metres.
Let C be the position of the cloud and C' be its reflection in the lake.
Then, CB = C'B.
Let PM be perpendicular from P on CB.
Then ∠ CPM = α and ∠ MPC' = β. Let CM = x.
Then CB = CM + MB = CM + MB = CM + PA = x + h.
In Δ CPM, we have
tan α = CM /PM
tan α = x/AB [Since PM = AB]
AB = x cot α ------------ (i)
In Δ PMC', we have
tan β = C'M /PM
tan β = x+2h/AB [Since C'M = C'B + BM = x + h + h]
AB = (x + 2h) cot β --------------- (ii)
From (i) and (ii),
x cot α = (x + 2h) cot β x (cot α - cot β)
= 2h cot β
x (1/tan α - 1/tan β ) = 2h /tan β
But height of the cloud = x + h
=
+ h
=