Question

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

• A. ${30}^o$
• B. ${60}^o$
• C. ${75}^o$
• D. ${90}^o$

Answer 1

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

Given $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$

$\Rightarrow \frac{a+b+2c}{(a+c)(b+c)}=\frac{3}{a+b+c}$

$\Rightarrow (a+b+2c)(a+b+c)=3(a+c)(b+c)$

$\Rightarrow (a+b{)}^2+3c(a+b)+2c^2=3(ab+ac+bc+c^2)$

$\Rightarrow a^2+b^2+2ab+3ac+3bc+2c^2=3ab+3bc+3ac+c^2$

$\Rightarrow a^2+b^2-c^2=ab$

$\therefore cosC=\frac{a^2+b^2-c^2}{2ab}=\frac{ab}{2ab}=\frac{1}{2}=\cos {60}^o$

$\Rightarrow C={60}^o$