In a Delta ABC, if (sin A + sin B + sin C)(si | vedclass

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(B) Given: $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B$ Using the identity $(x+y)(x-y) = x^2 - y^2$,we get: $(\sin A + \sin

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$\frac{\pi}{3}$

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(B) Given: $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B$ Using the identity $(x+y)(x-y) = x^2 - y^2$,we get: $(\sin A + \sin B)^2 - \sin^2 C = 3\sin A\sin B$ $\sin^2 A + \sin^2 B + 2\sin A\sin B - \sin^2 C = 3\sin A\sin B$ $\sin^2 A + \sin^2 B - \sin^2 C = \sin A\sin B$ By the Sine Rule,$\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$. Substituting these: $\frac{a^2}{4R^2} + \frac{b^2}{4R^2} - \frac{c^2}{4R^2} = \frac{a}{2R} \cdot \frac{b}{2R}$ $a^2 + b^2 - c^2 = ab$ Dividing by $2ab$: $\frac{a^2 + b^2 - c^2}{2ab} = \frac{ab}{2ab} = \frac{1}{2}$ By the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Therefore,$\cos C = \frac{1}{2} = \cos \frac{\pi}{3}$. Thus,$\angle C = \frac{\pi}{3}$.

In a $\Delta ABC,$ if $(\sin A + \sin B + \sin C)(\sin A + \sin B - \sin C) = 3\sin A\sin B,$ then the angle $C$ is equal to

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