Question:
Mark the correct alternative in each of the following:
In a $\triangle \mathrm{ABC}$, if $(c+a+b)(a+b-c)=a b$, then the measure of angle $C$ is
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $\frac{2 \pi}{3}$
(d) $\frac{\pi}{2}$
Solution:
Given: $(c+a+b)(a+b-c)=a b$
$\Rightarrow(a+b)^{2}-c^{2}=a b$
$\Rightarrow a^{2}+b^{2}+2 a b-c^{2}=a b$
$\Rightarrow a^{2}+b^{2}-c^{2}=-a b$
$\Rightarrow \frac{a^{2}+b^{2}-c^{2}}{2 a b}=-\frac{1}{2}$
$\Rightarrow \cos C=-\frac{1}{2}=\cos \frac{2 \pi}{3} \quad$ (Using cosine rule)
$\Rightarrow C=\frac{2 \pi}{3}$
Thus, the measure of angle $C$ is $\frac{2 \pi}{3}$.
Hence, the correct answer is option (c).