In a triangle ABC,if angle A = 60^{circ},then | vedclass

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(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.$\frac{1}{2} = \fr

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$1$

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(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$ $\Rightarrow bc = b^2+c^2-a^2$ $\Rightarrow b^2+c^2 = bc+a^2 \dots (i)$.Now,consider the expression $\frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)}$.$= \frac{ab+b^2+c^2+ac}{ac+a^2+bc+ab}$.Substitute $b^2+c^2 = bc+a^2$ from $(i)$:$= \frac{ab+(bc+a^2)+ac}{ac+a^2+bc+ab} = \frac{ab+bc+a^2+ac}{ab+bc+a^2+ac} = 1$.

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

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