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In a $\triangle A B C$, if $\angle A=60^{\circ}$, then $\frac{b}{c+a}+\frac{c}{a+b}=$
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Verified Answer
The correct answer is:
1
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,
$$
\begin{aligned}
& \frac{b}{c+a}+\frac{c}{a+b}=\frac{b(a+b)+c(c+a)}{(c+a)(a+b)} \\
& =\frac{a b+b^2+c^2+a c}{a c+a^2+b c+a b} \\
& =\frac{a b+b c+a^2+a c}{a c+a^2+b c+a b} \quad[\because \text { from Eq. (i) }] \\
& =1
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \cos 60^{\circ}=\frac{b^2+c^2-a^2}{2 b c} \Rightarrow \frac{1}{2}=\frac{b^2+c^2-a^2}{2 b c} \\
& \Rightarrow b c=b^2+c^2-a^2 \\
& \Rightarrow b^2+c^2=b c+a^2...(i)
\end{aligned}
$$
Now,
$$
\begin{aligned}
& \frac{b}{c+a}+\frac{c}{a+b}=\frac{b(a+b)+c(c+a)}{(c+a)(a+b)} \\
& =\frac{a b+b^2+c^2+a c}{a c+a^2+b c+a b} \\
& =\frac{a b+b c+a^2+a c}{a c+a^2+b c+a b} \quad[\because \text { from Eq. (i) }] \\
& =1
\end{aligned}
$$
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