In a $ riangle A B C$, if $\angle A=60^{\circ}$, then $\frac{b}{c+a}+\frac{c}{a+b}=$

Question

In a $	riangle A B C$, if $\angle A=60^{\circ}$, then $\frac{b}{c+a}+\frac{c}{a+b}=$

Aicharya · Accepted Answer

Apply cosine rule, $\cos A=\frac{b^2+c^2-a^2}{2 b c}$ Here, $A=60^{\circ}$ $ \begin{aligned} & \Rightarrow \cos 60^{\circ}=\frac{b^2+c^2-a^2}{2 b c} \Rightarrow \frac{1}{2}=\frac{b^2+c^2-a^2}{2 b c} \ & \Rightarrow b c=b^2+c^2-a^2 \ & \Rightarrow b^2+c^2=b c+a^2...(i) \end{aligned} $ Now, $ \begin{aligned} & \frac{b}{c+a}+\frac{c}{a+b}=\frac{b(a+b)+c(c+a)}{(c+a)(a+b)} \ & =\frac{a b+b^2+c^2+a c}{a c+a^2+b c+a b} \ & =\frac{a b+b c+a^2+a c}{a c+a^2+b c+a b} \quad[\because 	ext { from Eq. (i) }] \ & =1 \end{aligned} $

In a $\triangle A B C$, if $\angle A=60^{\circ}$, then $\frac{b}{c+a}+\frac{c}{a+b}=$

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