Question

If in a triangle ABC, $\angle C={60}^o,$ then $\frac{1}{a+c}+\frac{1}{b+c}=$
[IIT 1975]

• A. $\frac{1}{a+b+c}$
• B. $\frac{2}{a+b+c}$
• C. $\frac{3}{a+b+c}$
• D. None of these

Answer 1

The correct option is C

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

cos $C=\frac{\pi }{3}\Rightarrow a^2+b^2-c^2=ab$
$\Rightarrow b^2+bc+a^2+ac=ab+ac+bc+c^2$
$\Rightarrow$ b(b+c)+a(a+c)=(a+c)(b+c)
Divide by (a+c)(b+c) and add 2 on both sides
$\Rightarrow 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3\Rightarrow \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}.$