Question

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

• A. 90°
• B. 60°
• C. 45°
• D. 30°

Answer 1

The correct option is B

60°

Given that,

$\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}=1\Rightarrow a(c+a)+b(b+c)=(b+c)(c+a)\Rightarrow a^2+b^2-c^2=ab\therefore cosC=\frac{a^2+b^2-c^2}{2ab}\frac{ab}{2ab}\frac{1}{2}\Rightarrow C={60}^0$