If △ABC,8R2=a2+b2+c2, then the triangle ABC, is

We have,8R2=a2+b2+c2⇒8R2=4R2sin2⁡A+sin2⁡B+sin2⁡C⇒sin2⁡A+sin2⁡B+sin2⁡C=2⇒1−cos2⁡A+1−cos2⁡B+sin2⁡C=2⇒cos2⁡A−sin2⁡C+cos2⁡B=0⇒cos⁡(A+C)cos⁡(A−C)+cos2⁡B=0⇒−cos⁡B{cos⁡(A−C)−cos⁡B}=0⇒−cos⁡B{cos⁡(A−C)+cos⁡(A+C)}=0⇒−2cos⁡Acos⁡Bcos⁡C=0⇒cos⁡A=0 or, cos⁡B=0 or, cos⁡C=0⇒A=π2 or, B=π2 or, C=π2⇒ △ABC is a right angled triangle