Question

In triangle $ABC,$ 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. ${90}^{\circ }$

Answer 1

Answer: (C)

${60}^{\circ }$

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

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

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

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

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

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

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

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

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

$\cos C=\frac{1}{2}\Rightarrow \angle C={60}^{\circ }$