In a triangle the radius of the circumcircle is double the radius of the inscribed circle, then the triangle is

$R=2r$

We know that $r=4R\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}$

$\Rightarrow \sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2}=\frac{1}{8}$

$\Rightarrow \Sigma \cos A-1=\frac{1}{2}$

$\Rightarrow \cos A+\cos B+\cos C=\frac{3}{2}$

$\Rightarrow 2\cos \frac{A+B}{2}.\cos \frac{A-B}{2}+1-2{\sin }^2\frac{C}{2}=\frac{3}{2}$

$\Rightarrow 2\sin \frac{C}{2}.\cos \frac{A-B}{2}-2{\sin }^2\frac{C}{2}=\frac{1}{2}$

$\cos \frac{A-B}{2}\le 1$

$2\sin \frac{C}{2}.\cos \frac{A-B}{2}\le 2\sin \frac{C}{2}$

$\frac{1}{2}\le 2\sin \frac{C}{2}-2{\sin }^2\frac{C}{2}$

$4{\sin }^2\frac{C}{2}-4\sin \frac{C}{2}+1\le 0$

$4{\sin }^2\frac{C}{2}-4\sin \frac{C}{2}+1\le 0$

${(2\sin \frac{C}{2}-1)}^2\le 0$

$\sin \frac{C}{2}=\frac{1}{2}\Rightarrow C=60°$

$\text{and\hspace{0.17em}\hspace{0.17em}}\frac{A-B}{2}=0\Rightarrow A=B$