## anonymous one year ago sin^2A+sin^2B+sin^2C=2.......then show that the triangle is a right angled triangle

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1. amoodarya

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