Question

If, in $\Delta ABC$, $\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$ then the angle C =

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

Answer 1

Answer: (C)

${60}^{\circ }$

$\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$
$\frac{a+b+2c}{c^2+(a+b)c+ab}=\frac{3}{a+b+c}$
$(a+b+c{)}^2+c(a+b+c)=3c^2+3(a+b)c+3ab$
$a^2+b^2+c^2+2ab+2bc+2ac+ac+bc+c^2=3c^2+3ac+3bc+3ab$
$a^2+b^2+2c^2+3(ab+ac+bc)-ab=3c^2+3(ab+ac+bc)$
$a^2+b^2-ab=c^2$
$a^2+b^2-c^2=ab$
$\frac{a^2+b^2-c^2}{ab}=1$
$\frac{a^2+b^2-c^2}{2ab}=\frac{1}{2}$
$cosC=\frac{1}{2}$ .......................Cosine rule $\cos C=\frac{b^2+a^2-c^2}{2ab}$
$C={60}^0$