If cos^2 A + cos^2 C = sin^2 B,then Delta ABC | vedclass

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(B) Given: $\cos^2 A + \cos^2 C = \sin^2 B$Since $A + B + C = \pi$,we have $C = \pi - (A + B)$,so $\cos C = -\cos(A + B)$.Substituting this into the e

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Right-angled

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(B) Given: $\cos^2 A + \cos^2 C = \sin^2 B$Since $A + B + C = \pi$,we have $C = \pi - (A + B)$,so $\cos C = -\cos(A + B)$.Substituting this into the equation:$\cos^2 A + \cos^2(A + B) = \sin^2 B$$\cos^2 A + (\cos A \cos B - \sin A \sin B)^2 = \sin^2 B$Using the identity $\cos^2 A = 1 - \sin^2 A$ and simplifying,or testing $B = 90^\circ$:If $B = 90^\circ$,then $\sin^2 B = 1$.Then $\cos^2 A + \cos^2 C = \cos^2 A + \cos^2(90^\circ - A) = \cos^2 A + \sin^2 A = 1$.Since the condition holds for $B = 90^\circ$,the triangle is right-angled.

If $\cos^2 A + \cos^2 C = \sin^2 B$,then $\Delta ABC$ is

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