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A man on the top of a cliff $100 m$ high, observes the angles of depression of the two points on opposite sides of the cliff as $30 °$ and $60 °$ respectively, the distance between the two points is

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\frac{400}{\sqrt{3}}

\frac{200}{\sqrt{3}}

\frac{500}{\sqrt{3}}

\frac{100}{\sqrt{3}}

answer is A.

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Detailed Solution

In $Δ C A B, \tan 30 ° = \frac{100}{A C}$

$\frac{1}{\sqrt{3}} = \frac{100}{A C}$

$\Rightarrow A C = 100 \sqrt{3}$ _________(1)

In $Δ D A B, \tan 60 ° = \frac{100}{A D}$

$\sqrt{3} = \frac{100}{A D} \Rightarrow A D = \frac{100}{\sqrt{3}}$ ______(2)

$∴ C D = A C + A D = 100 \sqrt{3} + \frac{100}{\sqrt{3}}$

$= 100 (\sqrt{3} + \frac{1}{\sqrt{3}}) = \frac{400}{\sqrt{3}} m t$

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