In triangle ABC,if 8R^2 = a^2 + b^2 + c^2,the | vedclass

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(A) Given,$8R^2 = a^2 + b^2 + c^2$. Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these in the given equation:

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right angled triangle

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(A) Given,$8R^2 = a^2 + b^2 + c^2$. Using the sine rule,$a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$. Substituting these in the given equation: $8R^2 = (2R \sin A)^2 + (2R \sin B)^2 + (2R \sin C)^2$ $8R^2 = 4R^2 (\sin^2 A + \sin^2 B + \sin^2 C)$ $2 = \sin^2 A + \sin^2 B + \sin^2 C$ Using $\sin^2 	heta = 1 - \cos^2 	heta$: $2 = (1 - \cos^2 A) + (1 - \cos^2 B) + \sin^2 C$ $2 = 2 - \cos^2 A - \cos^2 B + \sin^2 C$ $\cos^2 A + \cos^2 B = \sin^2 C$ $\cos^2 A + \cos^2 B = 1 - \cos^2 C$ $\cos^2 A + \cos^2 B + \cos^2 C = 1$ This is a known identity for a right-angled triangle where one of the angles is $90^\circ$. Alternatively,$\cos^2 A + \cos^2 B + \cos^2 C = 1 - 2 \cos A \cos B \cos C = 1$ implies $\cos A \cos B \cos C = 0$. Thus,$A = 90^\circ$ or $B = 90^\circ$ or $C = 90^\circ$. Therefore,the triangle is a right-angled triangle.

In $\triangle ABC$,if $8R^2 = a^2 + b^2 + c^2$,then the triangle is a/an

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