Q. 13

# If the angle of Given:

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

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