Question

In$\Delta ABC$, If $\angle C={60}^{\circ }$ and $\frac{1}{a+c}+\frac{1}{(b+c)}=\frac{k}{a+b+c}$, then the value of $k$ is

• A. $1$
• B. $3$
• C. $4$
• D. $7$

Answer 1

Answer: (B)

$3$

Since $\cos C=\cos \frac{\pi }{3}=\frac{1}{2}$
$\therefore a^2+b^2-c^2=ab$
Adding $bc+ca$ on both sides,
$b^2+bc+a^2+ac=ab+bc+ca+c^2$
$\Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)$
$\Rightarrow \frac{b}{a+c}+\frac{a}{b+c}=1\Rightarrow (1+\frac{b}{a+c})+(1+\frac{a}{b+c})=3$
$\Rightarrow \frac{a+b+c}{a+c}+\frac{a+b+c}{b+c}=3$
$\Rightarrow \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\Rightarrow k=3$