What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?

$$\sin^2A+\sin^2B+\sin^2C$$
$$=1-(\cos^2A-\sin^2B)+1-\cos^2C$$
$$=2-\cos(A+B)\cos(A-B)-\cos C\cdot\cos C$$
$$=2-\cos(\pi-C)\cos(A-B)-\cos\{\pi-(A+B)\}\cdot\cos C$$
$$=2+\cos C\cos(A-B)+\cos(A+B)\cdot\cos C\text{ as }\cos(\pi-x)=-\cos x$$
$$=2+\cos C\{\cos(A-B)+\cos(A+B)\}$$
$$=2+2\cos A\cos B\cos C$$
$(1)$ If $2+2\cos A\cos B\cos C=2, \cos A\cos B\cos C=0$
$\implies $ at least one of $\cos A,\cos B,\cos C$ is $0$ which needs the respective angles $=\frac\pi2$
But we can have at most one angle $\ge \frac\pi2$ 
So, here we shall have exactly one angle $=\frac\pi2$
$(2)$ If $2+2\cos A\cos B\cos C>2, \cos A\cos B\cos C>0$
Either all of  $\cos A,\cos B,\cos C$ must be $>0\implies$ all the angles are acute
or exactly two cosine ratios $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible for a triangle
$(3)$ If $2+2\cos A\cos B\cos C<2, \cos A\cos B\cos C<0$
Either all the ratios  $<0$ which needs the respective angles $> \frac\pi2,$ which is impossible fro a triangle
or exactly one of the cosine ratios is $<0\implies $  the respective angle $> \frac\pi2,$ 
