Question

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

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

Answer 1

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

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