Question

In a $△ABC,$ if $A={60}^0$, then $\frac{b}{c+a}+\frac{c}{a+b}$ equals:

Answer 1

In a $△ABC,$ given $\angle A={60}^0$ then using cosine rule, we have
$\cos {60}^0=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow \frac{1}{2}=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow 1=\frac{b^2+c^2-a^2}{bc}$
$\Rightarrow bc=b^2+c^2-a^2$
$\Rightarrow b^2+c^2=a^2+bc$ ..........$\left(1\right)$
$\therefore$ Given expression$=\frac{b}{c+a}+\frac{c}{a+b}$
$=\frac{ab+b^2+c^2+ac}{(a+b)(c+a)}$
$=\frac{ab+a^2+bc+ac}{a^2+ac+ab+bc}$ by eqn$\left(1\right)$
$=1$