In triangle ABC, with usual notations, if (a | vedclass

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(B) Given the equation: $(a + b + c)(a + b - c) = ab$ This can be rewritten as: $((a + b) + c)((a + b) - c) = ab$ Using the identity $(x+y)(x-y) = x^2

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$\frac{2\pi}{3}$

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(B) Given the equation: $(a + b + c)(a + b - c) = ab$ This can be rewritten as: $((a + b) + c)((a + b) - c) = ab$ Using the identity $(x+y)(x-y) = x^2 - y^2$, we get: $(a + b)^2 - c^2 = ab$ Expanding the square: $a^2 + b^2 + 2ab - c^2 = ab$ Rearranging the terms: $a^2 + b^2 - c^2 = -ab$ Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Substitute $a^2 + b^2 - c^2 = -ab$ into the formula: $\cos C = \frac{-ab}{2ab} = -\frac{1}{2}$ Since $\cos C = -\frac{1}{2}$, the angle $C = \frac{2\pi}{3}$ or $120^\circ$.

In triangle $ABC$, with usual notations, if $(a + b + c)(a + b - c) = ab$, then the measure of angle $C$ is...

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