First slide

Trigonometric Identities

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. Then the product of the lengths of the line segments
$A_{0} A_{l}, A_{0} A_{2}, and A_{0} A_{4}$ is

Moderate

A

¾
B

$3 \sqrt{3}$
C

3
D

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

Solution

Let O be the center of the circle.
$∠ A_{0} {OA}_{1} = \frac{360^{\circ}}{6} = 60^{\circ}$
Thus, $A_{0} {OA}_{1}$ is an equilateral triangle. We get
$\begin{matrix} also A_{0} A_{1} = 1 \\ A_{0} A_{2} = A_{0} A_{4} \\ = 2 A_{0} D \\ = 2 ({OA}_{0} \sin 60^{\circ}) \\ = 2 (1) \frac{\sqrt{3}}{2} = \sqrt{3} \\ (A_{0} A_{1}) (A_{0} A_{2}) (A_{0} A_{4}) \\ = (1) (\sqrt{3}) (\sqrt{3}) = 3 \end{matrix}$

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