Let A_0 A_1 A_2 A_3 A_4 A_5 be a regular hexa | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a regular hexagon inscribed in a circle of unit radius,the side length is equal to the radius,so $A_0 A_1 = 1$.Since the interior angle of a re

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) In a regular hexagon inscribed in a circle of unit radius,the side length is equal to the radius,so $A_0 A_1 = 1$.Since the interior angle of a regular hexagon is $120^\circ$,in $	riangle A_0 A_1 A_2$,by the Law of Cosines:$A_0 A_2^2 = A_0 A_1^2 + A_1 A_2^2 - 2(A_0 A_1)(A_1 A_2) \cos(120^\circ)$$A_0 A_2^2 = 1^2 + 1^2 - 2(1)(1)(-\frac{1}{2}) = 1 + 1 + 1 = 3$Thus,$A_0 A_2 = \sqrt{3}$.By symmetry,$A_0 A_4 = A_0 A_2 = \sqrt{3}$.The product of the lengths is $A_0 A_1 	imes A_0 A_2 	imes A_0 A_4 = 1 	imes \sqrt{3} 	imes \sqrt{3} = 3$.

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 $A_0 A_1$,$A_0 A_2$,and $A_0 A_4$ is

Similar Questions

The value of $\cos^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{4} + \cos^2 \frac{5\pi}{12}$ is

If $\alpha, \beta, \gamma \in \left(0, \frac{\pi}{2}\right)$,then $\frac{\sin(\alpha + \beta + \gamma)}{\sin \alpha + \sin \beta + \sin \gamma}$ is

If $\alpha = \frac{\sin^3 x}{\cos^2 x}$,$\beta = \frac{\cos^3 x}{\sin^2 x}$ and $\sin x + \cos x = k$,then $\alpha \sin x + \beta \cos x + 3 = $

If $\cos (x-y), \cos x, \cos (x+y)$ are three distinct numbers which are in harmonic progression and $\cos x \neq \cos y$,then $1+\cos y$ is equal to

Evaluate: $\sin \frac{\pi}{12} \sin \frac{2 \pi}{12} \sin \frac{3 \pi}{12} \sin \frac{4 \pi}{12} \sin \frac{5 \pi}{12} \sin \frac{6 \pi}{12}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module