Question

The value of $A_i{A}_j$ where $j=1,2,3,...n$ is

• A. $2R\sin \left[(j-1)\frac{\pi }{2n}\right]$
• B. $R\sin \left[(j-1)\frac{2\pi }{n}\right]$
• C. $2R\sin \left[(j-1)\frac{\pi }{n}\right]$
• D. $R\sin \left[(j-1)\frac{\pi }{n}\right]$

Answer 1

The correct option is C $2R\sin \left[(j-1)\frac{\pi }{n}\right]$

$\angle A_{1}O{A}_2=\frac{2\pi }{n}$
$\therefore A_1{A}_2=\sqrt{R^2+R^2-2R.R\cos \left(\frac{2\pi }{n}\right)}$
Since $cos2\theta =1-2{\sin }^2\theta$ we have $\cos \left(\frac{2\pi }{n}\right)=1-2{\sin }^2\frac{\pi }{n}$
$\Rightarrow A_1{A}_2=\sqrt{2R^2-2R^2(1-2{\sin }^2\frac{\pi }{n})}$
$=\sqrt{4R^2{\sin }^2\left(\frac{\pi }{n}\right)}$
$=2R\sin \frac{\pi }{n}$ ...........$\left(1\right)$
Consider $A_1{A}_3=\sqrt{R^2+R^2-2R.R\cos \left(\frac{4\pi }{n}\right)}$
$=\sqrt{2R^2-2R^2(1-2{\sin }^2\frac{4\pi }{n})}$
$=\sqrt{4R^2{\sin }^2\left(\frac{4\pi }{n}\right)}$
$=2R\sin \frac{2\pi }{n}$ ...........$\left(2\right)$
Continuing like this
$A_1{A}_4=2R\sin \left(\frac{3\pi }{n}\right)$ and so on.
$\frac{\pi }{n},\frac{2\pi }{n},\frac{3\pi }{n}...$ are in A.P
$\therefore {(j-1)}^{th}$ term is $=\frac{\pi }{n}+(j-2)\frac{\pi }{n}=(j-1)\frac{\pi }{n}$
$\therefore A_i{A}_j=2R\sin (j-1)\frac{\pi }{n}$ where $j=1,2,3,...n$