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Q.

A man is at a height of 100 m. He sees a car which makes an angle of $\frac{π}{6}$ with horizontal with man's position. If the car moves to .a point where angle is $\frac{π}{3}$ , what is the distance moved by it ?

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a

$(\frac{100}{\sqrt{3}}) m$

b

$200 \sqrt{3} m$

c

$(\frac{200}{\sqrt{3}}) m$

d

$(\frac{150}{\sqrt{3}}) m$

answer is C.

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Detailed Solution

See figure

Question Image

The distance moved is y.
From $△ ABC$
$x = \frac{100}{\tan 60^{\circ}} = \frac{100}{\sqrt{3}}$
From $△ ABD$
$\tan 30^{\circ} = \frac{100}{x + y}$
or $(x + y) = \frac{100}{\tan 30^{\circ}} = \frac{100 \times \sqrt{3}}{1}$
Now y=(x+ y)-x
$= 100 \sqrt{3} - \frac{100}{\sqrt{3}} = \frac{200}{\sqrt{3}}$

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