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If the angle of elevation of an object from a point $100 m$ above a lake is found to be $30 °$ and the angle of depression of its image in the lake is $45 °$ , then the height of the object above the lake is

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100 (2 - \sqrt{3}) m

100 (2 + \sqrt{3}) m

100 (\sqrt{3} - 1) m

1000 (\sqrt{3} + 1) m

answer is B.

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Detailed Solution

It is given that the angle of elevation of an object from a point

100 m

above a lake is

30 ° .

The angle of depression of its image in the lake is

45 °

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

Let

AB

the surface of the lake

O

be the point of observation

P

be the object and

P^{'}

be its image.
Let the height of object

P

above lake be

x .

So, it will be below the lake by the same distance as they are mirror images.

EP = P^{'} E = x

OC = DE = 100 m

and

PD = P' E = DE

\Rightarrow PD = x - 100

P' D = x + 100

△ OPD,

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

\tan 30 ° = \frac{PD}{OD}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{x - 100}{OD}

∵ tan 30 ° = 13

\Rightarrow OD = \sqrt{3} (x - 100) m

△ O P^{'} D,

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

\tan 45 ° = \frac{P' D}{OD}

\Rightarrow 1 = \frac{P' D}{OD}

∵ tan 45 ° = 1

\Rightarrow OD = P' D

Substituting values of

OD

and

P' D

\Rightarrow \sqrt{3} (x - 100) = x + 100

\Rightarrow x = \frac{100 (\sqrt{3} + 1)}{\sqrt{3} - 1}

Now rationalizing the denominator we get,

\Rightarrow x = \frac{100 (\sqrt{3} + 1)}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}

\Rightarrow x = \frac{100 (3 + 1 + 2 \sqrt{3})}{3 - 1}

\Rightarrow x = 100 (2 + \sqrt{3})

Hence, the height of the object above the lake is

100 (2 + \sqrt{3}) m .

So, the correct option is 2.

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