2. Let $a, b, c$ be the lengths of the sides opposite to the angles $\angle A, \angle B, \angle C$ of $\triangle A B C$ respectively, and $\angle A=60^{\circ}$. Then the value of $\frac{c}{a+b}+\frac{b}{a+c}$ is $\qquad$
2. 1

Solution By the cosine rule $a^{2}=b^{2}+c^{2}-2 b c \cos A=b^{2}+c^{2}-b c$, thus
$$
\begin{aligned}
\frac{c}{a+b}+\frac{b}{a+c} & =\frac{a c+c^{2}+a b+b^{2}}{a^{2}+a b+a c+b c} \\
& =\frac{a b+a c+b^{2}+c^{2}}{a b+a c+b^{2}+c^{2}}=1 .
\end{aligned}
$$