Question

solve this question

Answer 1

@ hey bro...@

X = 10 m

Hope it will be your answer!!

Thanks!!

Answer 2

Let A be a point h meters above the lake 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's

=> BC = X

so

=> BF = (x+h)

=> BF = DF = (x+h) meters
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Now,

In △BAC we have :-

$= > cosec \alpha = \frac{AB}{x} \\ \\ = > AB = x.cosec \alpha \: \: \: \: . \\ \\ = > AC = x.cot \alpha \: \: \: ....(EQ _{1})$
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In△ACD we have :-

$= > cot \beta = \frac{AC}{2h + x} \\ \\ = > AC = (2h + x).cot \beta \: \: \: ..(EQ _{2})$
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By eq(1) and eq(2)

$= > x.cot \alpha = (2h + x).cot \beta \\ \\ = > x = \frac{2h.cot \beta }{cot \alpha - cot \beta }$
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plug the value of x

then,

$= > AB = \frac{cosec \alpha .2h.cot \beta }{cot \alpha - cot \beta } \\ \\ = > AB = \frac{2h.sec \alpha }{tan \beta - tan \alpha }$
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Now 2nd part of the question :-

$= > \alpha = 30 \\ \\ and \\ \\ = > \beta = 45$
Then

$= > x = \frac{2h.cot \beta }{cot \alpha - cot \beta } \\ \\ = > x = \frac{2h.cot45}{cot30 - cot45} \\ \\ = > x = \frac{2h \times 1}{ \sqrt{3} - 1 } \\ \\ = > x = \frac{2h}{ \sqrt{3} - 1 } \times \frac{ \sqrt{3} + 1 }{ \sqrt{3} + 1} \\ \\ = > x = h( \sqrt{3} + 1)$
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