In triangle ABC,if a, b, c are its sides and | vedclass

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(A) In $	riangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Given $\angle C = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2} = \fra

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(A) In $	riangle ABC$,by the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.Given $\angle C = 60^{\circ}$,so $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.This implies $ab = a^2+b^2-c^2$,or $c^2 = a^2+b^2-ab$.Now,consider the expression $E = \frac{a}{b+c} + \frac{b}{c+a} = \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} = \frac{ac+a^2+b^2+bc}{bc+ab+c^2+ac}$.Substituting $a^2+b^2 = c^2+ab$ into the numerator:$E = \frac{ac + (c^2+ab) + bc}{bc+ab+c^2+ac} = \frac{ac+c^2+ab+bc}{ac+ab+c^2+bc} = 1$.

In $\triangle ABC$,if $a, b, c$ are its sides and $\angle C = 60^{\circ}$,find the value of $\frac{a}{b+c} + \frac{b}{c+a}$.

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