$\triangle A B C$, if $a, b, c$ are its sides and $\angle C=60^{\circ}$, find the value of…
$\triangle A B C$, if $a, b, c$ are its sides and $\angle C=60^{\circ}$, find the value of $\frac{a}{b+c}+\frac{b}{c+a}$
1
0
$\frac{\sqrt{3}}{2}$
$\frac{1}{2}$
Solution
In a $\triangle A B C$, it is given $\angle C=60^{\circ}$, then
$
\begin{aligned}
& \cos 60^{\circ}=\frac{a^2+b^2-c^2}{2 a b} \\
& \Rightarrow a b=a^2+b^2-c^2 \Rightarrow c^2=a^2+b^2-a b
\end{aligned}
$
Now, $\frac{a}{b+c}+\frac{b}{c+a}=\frac{a c+a^2+b^2+b c}{b c+a b+c^2+a c}$
$
\begin{aligned}
& =\frac{a c+c^2+a b+b c}{a c+a b+c^2+b c} \quad\left\{\because a^2+b^2=c^2+a b\right\} \\
& =1
\end{aligned}
$