Question

Let $A_0,A_1,A_2,A_3,A_4$ and $A_5$ be the consecutive vertices of a regular hexagon inscribed in a unit circle. The product of the lengths of $A_0{A}_1,$ $A_0{A}_2$ and $A_0{A}_4$ is

• A. $\frac{3}{4}$
• B. $3\sqrt{3}$
• C. $3$
• D. $\frac{3\sqrt{3}}{2}$

Answer 1

Answer: (C)

$3$

In regular hexagon $A_0{A}_1{A}_2{A}_3{A}_4{A}_5$
$O{A}_0=O{A}_1=O{A}_2=O{A}_3=O{A}_4=O{A}_5=1\&\angle A_{0}O{A}_1=\angle A_{1}O{A}_2=\angle A_{2}O{A}_3=\angle A_{3}O{A}_4=\angle A_{4}O{A}_5=\angle A_{5}O{A}_0=\frac{\pi }{3}$
In equilateral triangle $A_{0}O{A}_1$
$\therefore O{A}_1=A_0{A}_1=1$
In Isosceles triangle $A_{0}O{A}_2$
$\angle {OA}_0{A}_2=\angle {OA}_2{A}_0=\frac{\pi }{6}\&\angle A_{0}O{A}_2=\frac{2\pi }{3}$
From Sine rule:
$\frac{O{A}_0}{\sin \angle {OA}_2{A}_0}=\frac{A_0{A}_2}{\sin \angle A_{0}O{A}_2}\Rightarrow \frac{1}{\sin \frac{\pi }{6}}=\frac{A_0{A}_2}{\sin \frac{2\pi }{3}}\Rightarrow A_0{A}_2=\sqrt{3}\therefore A_0{A}_4=A_0{A}_2=\sqrt{3}\Rightarrow {A_0{A}_{1}A}_0{A}_4{A}_0{A}_2=3$
Ans: C