Question

Mark the correct alternative in each of the following:

In a ∆ABC, if $\left(c+a+b\right)\left(a+b-c\right)=ab$, then the measure of angle C is

(a) $\frac{\mathrm{\pi }}{3}$ (b) $\frac{\mathrm{\pi }}{6}$ (c) $\frac{2\mathrm{\pi }}{3}$ (d) $\frac{\mathrm{\pi }}{2}$

Answer 1

Given: $\left(c+a+b\right)\left(a+b-c\right)=ab$

$$\begin{gathered} \Rightarrow {\left(a+b\right)}^2-c^2=ab \\ \Rightarrow a^2+b^2+2ab-c^2=ab \\ \Rightarrow a^2+b^2-c^2=-ab \\ \Rightarrow \frac{a^2+b^2-c^2}{2ab}=-\frac{1}{2} \end{gathered}$$
$$\begin{gathered} \Rightarrow \cos C=-\frac{1}{2}=\cos \frac{2\mathrm{\pi }}{3}\left(\mathrm{Using}\mathrm{cosine}\mathrm{rule}\right) \\ \Rightarrow C=\frac{2\mathrm{\pi }}{3} \end{gathered}$$
Thus, the measure of angle C is $\frac{2\mathrm{\pi }}{3}$.

Hence, the correct answer is option (c).