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From a point $a$ meter above the lake, the angle of elevation of a cloud is $α$ and the angle of depression of its reflection is $β$ .The height of the cloud is

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\frac{asin (α + β)}{\sin (α - β)} m

\frac{asin (α + β)}{\sin (β - α)} m

\frac{asin (β - α)}{\sin (α + β)} m

None of these

answer is B.

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Detailed Solution

It is given that the angle of elevation of a cloud is

α

from a point which is

a

meter above the lake, and the angle of depression of its reflection is

β

on the lake.
According to the given data the figure has been drawn below.
Question Image

Let

OQ

be the surface of lake ,

P

be the point

a

metre above the lake ,

C

is the point where cloud is present at height

H

and

C^{'}

is reflection of cloud under the lake.
In

△ PAC,

tan θ = P e r p e n d i c u l a r b a s e

tanα = \frac{CA}{PA}

= \frac{H - a}{x}

\Rightarrow x = (H - a) cotα

(tanθ = \frac{1}{cotθ})

....(i)
In

△ PA C^{'},

tan θ = P e r p e n d i c u l a r b a s e

tanβ = \frac{AC'}{PA}

= \frac{H + a}{x}

\Rightarrow x = (H + a) cotβ

(tanθ = \frac{1}{cotθ})

....(ii) From equations

(i)

and

(ii)

, we get

(H + a) cotβ = (H - a) cotα

Hcotβ + acotβ = Hcotα - acotα

Hcotα - Hcotβ = acotα + acotβ

\Rightarrow H = \frac{a (cotα + cotβ)}{cotα - cotβ}

\Rightarrow \frac{a (cotα + cotβ)}{cotα - cotβ}

\Rightarrow \frac{a (\frac{cosα}{sinα} + {\frac{cosβ}{sinβ}}_{})}{\frac{cosα}{sinα} - \frac{cosβ}{sinβ}}

(cotθ = \frac{cosθ}{sinθ})

\Rightarrow \frac{a (cosαsinβ + cosβsinα)}{cosαsinβ - cosβsinα}

\Rightarrow \frac{asin (α + β)}{\sin (β - α)}

∵ sin (A + B) = sin A cos B + cos A sin B

= \frac{asin (α + β)}{\sin (β - α)}

.
So the height of the cloud is

\frac{asin (α + β)}{\sin (β - α)}

.
So the correct option is 2.

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