Question

sin^2A+sin^2B+sin^2C=2.......then show that the triangle is a right angled triangle

Answer 1

\[\sin^2(A)=\frac{1-\cos(2A)}{2}\\sin^2(B)=\frac{1-\cos(2B)}{2}\\sin^2(C)=\frac{1-\cos(2C)}{2}\\\frac{3-(\cos(2A)+\cos(2B)+\cos(2C)}{2}=2\\cos(2A)+\cos(2B)+\cos(2C)=-1\\cos(2A)+2 \cos(\frac{2B+2C}{2})\cos(\frac{2B-2C}{2})=-1\]
then put B+C=180-A
and simplify