In a triangle ABC,with usual notations,(a+b+c | vedclass

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(C) Given the equation $(a+b+c)(a+b-c) = 3ab$. Expanding the left side,we get $((a+b)+c)((a+b)-c) = 3ab$. This simplifies to $(a+b)^2 - c^2 = 3ab$. Ex

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$\frac{\pi}{3}$

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(C) Given the equation $(a+b+c)(a+b-c) = 3ab$. Expanding the left side,we get $((a+b)+c)((a+b)-c) = 3ab$. This simplifies to $(a+b)^2 - c^2 = 3ab$. Expanding $(a+b)^2$,we have $a^2 + b^2 + 2ab - c^2 = 3ab$. Rearranging the terms,we get $a^2 + b^2 - c^2 = ab$. Dividing both sides by $2ab$,we get $\frac{a^2 + b^2 - c^2}{2ab} = \frac{ab}{2ab} = \frac{1}{2}$. By the Cosine Rule,$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$. Therefore,$\cos C = \frac{1}{2}$. Since $\cos C = \frac{1}{2}$,we have $\angle C = \frac{\pi}{3}$ or $60^\circ$.

In a triangle $ABC$,with usual notations,$(a+b+c)(a+b-c)=3ab$,then $\angle C=$

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