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A flagstaff $10 m$ high stands at the centre of an equilateral triangle which is horizontal at the top of the flagstaff. Each side subtends at an angle of $60^{o}$ . What is the length of each side of the triangle?

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\frac{6 \sqrt{3}}{3}

\frac{4 \sqrt{6}}{3}

\frac{20}{\sqrt{3}}

\frac{6 \sqrt{5}}{3}

answer is C.

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Detailed Solution

It is given that a flagstaff

10 m

high stands at the centre of an equilateral triangle which is horizontal at the top of the flagstaff.
Each side subtends at an angle of

60^{o}

.
Let ABC be the equilateral triangle, with AD

⊥

BC.

Considering right angle

△ ADC

, we have,

\tan 60^{o} = \frac{AD}{DC}

⇒

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

⇒

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

m
⇒

DC = \frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}

m
⇒

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

m
⇒

BC = 2 \times DC

⇒

BC = 2 \times \frac{10 \sqrt{3}}{3}

m
⇒

BC = \frac{20 \sqrt{3}}{3}

∴

The length of each side of the triangle is

\frac{20}{\sqrt{3}} m

.
Hence, the correct option is 3.

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