Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$, then $\angle C$ is equal to :

• A. ${30}^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${75}^{\circ }$

Answer 1

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

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

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

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

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

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

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

$\Rightarrow C={60}^o$