ISI2018-DCG-22

Question

Let the sides opposite to the angles $A,B,C$ in a triangle $ABC$ be represented by $a,b,c$ respectively. If $(c+a+b)(a+b-c)=ab,$ then the angle $C$ is

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{2\pi}{3}$

Comments

$\mathbf D$ is the answer.

Answer 1

$\underline{\mathbf{Answer: D}}$

$\underline{\mathbf{Explanation:}}$

This is a very easy example of $\mathbf {SOT}$.

$\mathrm{(a+b+c) (a+b-c) = ab\;\;\;[Given]\\(a+b)^2 -c^2 = ab\;\;\;[Using\; Identity \;a^2-b^2 = (a+b) (a-b)] \\ a^2 + b^2 +2ab -c^2 = ab \\ \because a^2 + 2ab + b^2 = (a+b)^2 \\ \therefore a^2 + b^2 -c^2 = ab – 2ab \\ \Rightarrow a^2 + b^2 -c^2 = -ab \\ \Rightarrow \frac{a^2+b^2 -c^2}{ab} = -1}$

Dividing both sides by $2$, we get:

$\mathrm{\frac{a^2+b^2 -c^2}{2ab} = \frac{-1}{2}}$

$\mathrm{\Rightarrow \cos C = \frac{-1}{2} \\ \Rightarrow \cos C = \cos120^0  = \cos\frac{2\pi}{3} \\ \Rightarrow C = \frac{2\pi}{3}}$

$\therefore \mathbf D$ is the correct answer.