Question

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Answer 1

Answer:

Step-by-step explanation:

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.