If $\mathrm{△}ABC,8R^2=a^2+b^2+c^2$, then the triangle $ABC,$ is

We have,

$\begin{array}{l}8R^2=a^2+b^2+c^2\\ \Rightarrow 8R^2=4R^2\left({\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C\right)\\ \Rightarrow {\sin }^{2}A+{\sin }^{2}B+{\sin }^{2}C=2\\ \Rightarrow 1-{\cos }^{2}A+1-{\cos }^{2}B+{\sin }^{2}C=2\\ \Rightarrow \left({\cos }^{2}A-{\sin }^{2}C\right)+{\cos }^{2}B=0\\ \Rightarrow \cos (A+C)\cos (A-C)+{\cos }^{2}B=0\\ \Rightarrow -\cos B\{\cos (A-C)-\cos B\}=0\\ \Rightarrow -\cos B\{\cos (A-C)+\cos (A+C\left)\right\}=0\\ \Rightarrow -2\cos A\cos B\cos C=0\\ \Rightarrow \cos A=0\text{ or, }\cos B=0\text{ or, }\cos C=0\\ \Rightarrow A=\frac{\pi }{2}\text{ or, }B=\frac{\pi }{2}\text{ or, }C=\frac{\pi }{2}\end{array}$

$\Rightarrow \mathrm{△}ABC$ is a right angled triangle