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Trigonometric Ratios Using Right Angled Triangle

Let A0A1A2A...

Question

Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius.The product of the length of the line segments $A_{0} A_{1}, A_{0} A_{2}$ and $A_{0} A_{4}$ is

A

\frac{3}{4}

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B

3 \sqrt{3}

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C

3

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D

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

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Solution

The correct option is A $3$
$A_{0} A_{1} = 2.1. cos \frac{π}{3} = 2.1. \frac{1}{2} = 1 = A_{1} A_{2}$
$cos \frac{2 π}{3} = \frac{{(A_{0} A_{1})}^{2} + {(A_{1} A_{2})}^{2} - {(A_{0} A_{2})}^{2}}{2 A_{0} A_{1} . A_{1} A_{2}}$
$\Rightarrow cos (π - \frac{π}{3}) = \frac{1 + 1 - {(A_{0} A_{2})}^{2}}{2.1.1}$
$\Rightarrow - cos \frac{π}{3} = \frac{2 - {(A_{0} A_{2})}^{2}}{2}$
$\Rightarrow \frac{- 1}{2} = \frac{2 - {(A_{0} A_{2})}^{2}}{2}$
$\Rightarrow - 1 = 2 - {(A_{0} A_{2})}^{2}$
$\Rightarrow {(A_{0} A_{2})}^{2} = 3$
$∴ A_{0} A_{2} = \sqrt{3} = A_{0} A_{4}$
$∴ A_{0} A_{1} \times A_{0} A_{2} \times A_{0} A_{4} = 1 \times \sqrt{3} \times \sqrt{3} = 3$

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