Question

If $A=\frac{\pi }{7},B=\frac{2\pi }{7}$ and $C=\frac{4\pi }{7},$ then in $△ABC$ the value of $\frac{a^2+b^2+c^2}{R^2}$ is equals to

Answer 1

Using sine rule, $a=2R\sin A,b=2R\sin B,c=2R\sin C$
$\Rightarrow a^2+b^2+c^2=2R^2(1-\cos 2A+1-\cos 2B+1-\cos 2C)=2R^2[3-\cos \frac{2\pi }{7}-\cos \frac{4\pi }{7}-\cos \frac{8\pi }{7}]$
Now, using 7th roots of unity,we have $1+2(\cos \frac{2\pi }{7}+\cos \frac{4\pi }{7}+\cos \frac{6\pi }{7})=0$
$\Rightarrow a^2+b^2+c^2=2R^2[3-(-\frac{1}{2})]=7R^2[\because \frac{8\pi }{7}=2\pi -\frac{6\pi }{7}]\therefore \frac{a^2+b^2+c^2}{R^2}=7$