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Q.

The base of a cliff is circular. From the extremities of a diameter of the base, angles of elevation of the top of the cliff are $30^{\circ}$ and $60^{\circ}$ . If the height of the cliff be $500 m$ , then the diameter of the base of the cliff is

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a

\frac{2000}{\sqrt{3}} m

b

\frac{1000}{\sqrt{3}} m

c

\frac{2000}{\sqrt{2}} m

d

1000 \sqrt{3} m

answer is A.

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Detailed Solution

It is given that the angles of elevation of the top of the cliff are

30^{\circ}

and

60^{\circ}

The height of the cliff be

500 m

.
We have to find the diameter of the base of the cliff.
We use

tanθ = \frac{perpendicular}{base} .

Question Image

In

△ AEC

,

\tan 60^{o} = \frac{500}{d_{1}}

\Rightarrow \sqrt{3} = \frac{500}{d_{1}}

[since

\tan 60^{o} = \sqrt{3}

]

\Rightarrow d_{1} = \frac{500}{\sqrt{3}} m

Similarly, in

△ BEC

,

\tan 30^{o} = \frac{500}{d_{2}}

\Rightarrow \frac{1}{\sqrt{3}} = \frac{500}{d_{2}}

[since

\tan 30^{o} = \frac{1}{\sqrt{3}}

]

\Rightarrow d_{2} = 500 \sqrt{3} m

∴

Required diameter

= d_{1} + d_{2}

\Rightarrow diameter = \frac{500}{\sqrt{3}} + 500 \sqrt{3}

\Rightarrow diameter = \frac{2000}{\sqrt{3}} m

Thus, the diameter of the base of the cliff is

\frac{2000}{\sqrt{3}} m

.
Therefore, the correct option is 1.

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