In Delta ABC, if 8{R^2} = {a^2} + {b^2} + {c^ | vedclass

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(A) We know that $a = 2R \sin A,$ $b = 2R \sin B,$ and $c = 2R \sin C.$Substituting these in the given equation $8R^2 = a^2 + b^2 + c^2,$$8R^2 = (2R \

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Right angled

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(A) We know that $a = 2R \sin A,$ $b = 2R \sin B,$ and $c = 2R \sin C.$Substituting these in the given equation $8R^2 = a^2 + b^2 + c^2,$$8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$$8R^2 = 4R^2(\sin^2 A + \sin^2 B + \sin^2 C)$$2 = \sin^2 A + \sin^2 B + \sin^2 C$Using $\sin^2 	heta = 1 - \cos^2 	heta,$$2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$$2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$$\cos^2 A + \cos^2 B = \sin^2 C = 1 - \cos^2 C$$\cos^2 A + \cos^2 B + \cos^2 C = 1$Using the identity $\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C,$$1 = 1 - 2 \cos A \cos B \cos C$$2 \cos A \cos B \cos C = 0$This implies $\cos A = 0$ or $\cos B = 0$ or $\cos C = 0.$Therefore,$A = \frac{\pi}{2}$ or $B = \frac{\pi}{2}$ or $C = \frac{\pi}{2}.$Thus,the triangle is a right-angled triangle.

In $\Delta ABC,$ if $8{R^2} = {a^2} + {b^2} + {c^2},$ then the triangle is

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