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A vertical pole of height $10$ metres stand at one corner of a rectangular field. The angle of elevation of its top from the farthest corner is $30^{\circ}$ , while that from another corner is $60^{\circ}$ . Find the area (in $m^{2})$ of the rectangular field.

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\frac{200 \sqrt{2}}{3}

m^{2}

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

m^{2}

\frac{200 \sqrt{2}}{\sqrt{3}}

m^{2}

\frac{400 \sqrt{2}}{\sqrt{3}}

m^{2}

answer is A.

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Detailed Solution

Given that a vertical pole of height

10

metres stand at one corner of a rectangular field. The angle of elevation of its top from the farthest corner is

30^{\circ}

, while that from another corner is

60^{\circ}

In right triangle AEC,

\tan 30^{\circ} = \frac{10}{AC}

\Rightarrow \frac{10}{AC} = \frac{1}{\sqrt{3}}

\Rightarrow AC = 10 \sqrt{3} m

From the figure, the length of the rectangle is AB and breadth is BC.
Using pythagoras theorem in triangle ABC, where AC is diagonal,

∴ \sqrt{l^{2} + b^{2}} = AC = 10 \sqrt{3} m

-----------(i)
In triangle DCE,

\tan 60^{\circ} = \frac{10}{DC} = \sqrt{3}

\Rightarrow DC = \frac{10}{\sqrt{3}} = l

\Rightarrow l^{2} = \frac{100}{3}

----------(ii)

\Rightarrow b^{2} = {AC}^{2} - l^{2}

\Rightarrow b^{2} = 300 - \frac{100}{3}

[From (i) and (ii)]

\Rightarrow b^{2} = \frac{800}{3}

\Rightarrow b = \frac{\sqrt{800}}{\sqrt{3}}

Area = l \times b

\Rightarrow Area = \frac{\sqrt{800}}{\sqrt{3}} \times \frac{10}{\sqrt{3}}

\Rightarrow Area = \frac{10 \times 2 \times 10 \sqrt{2}}{3}

\Rightarrow Area = \frac{200 \sqrt{2}}{3}

So, the area of field is

\frac{200 \sqrt{2}}{3} m^{2}

.
Hence, the correct option is 1.

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