Question

In triangle $ABC,$ $If\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ then angle $C$ is equal to
[Note: All symbols used have usual meaning in triangle ABC.]

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${120}^o$

Answer 1

The correct option is C ${60}^{\circ }$

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

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

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

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

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

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

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

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

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

$\Rightarrow \cos C=\frac{1}{2}$

$\therefore C={60}^{\circ }$