Question

In $\Delta ABC$, if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$, then $\angle C$ is equal to

• A. ${90}^0$
• B. ${60}^0$
• C. ${45}^0$
• D. ${30}^0$

Answer 1

The correct option is B ${60}^0$

Given that, $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$
$\Rightarrow \frac{a+b+2c}{bc+ab+c^2+ac}=\frac{3}{a+b+c}$
$\Rightarrow (a+b+2c)(a+b+c)=3ab+3bc+3ac+3c^2$
$\Rightarrow a^2+b^2+2ab+3ac+3bc+2c^2=3ab+3bc+3ac+3c^2$
$\Rightarrow a^2+b^2-c^2=ab$
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}$
$\therefore \cos C=\frac{1}{2}\Rightarrow \angle C={60}^0$