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Parametric Equation of a Circle

Let A 0, A 1,...

Question

Let $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ be a regular hexagon inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments $A_{0} A_{1}, A_{0} A_{2}, A_{0} A_{4}$ is

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Solution

The correct option is C

Let O be the centre of the circle of unit radius and the coordinates of $A_{0}$ be (1,0)
Since each side of the regular hexagon makes an angle of $60^{\circ}$ at the centre O.
Coordinates of $A_{1} a r e (c o s 60^{\circ}, s i n 60^{\circ}) = (\frac{1}{2}, \frac{\sqrt{3}}{2})$
$A_{2} a r e (c o s 120^{\circ}, s i n 120^{\circ}) = (- \frac{1}{2}, \frac{\sqrt{3}}{2})$
$A_{3} a r e (- 1, 0)$
$A_{4} a r e (- \frac{1}{2}, - \frac{\sqrt{3}}{2}) a n d A_{5} a r e (\frac{1}{2}, - \frac{\sqrt{3}}{2})$
Now $A_{0} A_{1} = \sqrt{{(1 - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$
$A_{0} A_{2} = \sqrt{{(1 + \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} = A_{0} A_{4}$
So that $(A_{0} A_{1})$ $(A_{0} A_{2})$ $(A_{0} A_{4}) = 3$

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Let A0,A1,A2,A3,A4,A5 be a regular hexagon inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments A0A1,A0A2,A0A4 is

Let $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}, A_{5}$ be a regular hexagon inscribed in a unit circle with centre at the origin. Then the product of the lengths of the line segments $A_{0} A_{1}, A_{0} A_{2}, A_{0} A_{4}$ is