In Delta ABC,if sin^{2} A + sin^{2} B + sin^{ | vedclass

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(B) Given $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$.Using $\sin^{2} 	heta = 1 - \cos^{2} 	heta$,we have:$(1 - \cos^{2} A) + \sin^{2} B + \sin^{2} C

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Right-angled

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(B) Given $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$.Using $\sin^{2} 	heta = 1 - \cos^{2} 	heta$,we have:$(1 - \cos^{2} A) + \sin^{2} B + \sin^{2} C = 2$$\sin^{2} B + \sin^{2} C - \cos^{2} A = 1$Using the identity $\sin^{2} B + \sin^{2} C = 1 - \cos(B+C)\cos(B-C)$ and $\cos A = -\cos(B+C)$:$1 - \cos(B+C)\cos(B-C) - \cos^{2} A = 1$$-\cos(B+C)\cos(B-C) - \cos^{2} A = 0$$\cos A \cos(B-C) - \cos^{2} A = 0$$\cos A [\cos(B-C) - \cos A] = 0$$\cos A [\cos(B-C) + \cos(B+C)] = 0$$\cos A [2 \cos B \cos C] = 0$This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0$.Thus,one of the angles must be $90^{\circ}$,which means the triangle is right-angled.

In $\Delta ABC$,if $\sin^{2} A + \sin^{2} B + \sin^{2} C = 2$,then the triangle is:

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