In triangle ABC,if frac{1}{b+c} + frac{1}{c+a | vedclass

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(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(

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$60$

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(B) Given the equation: $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Taking the common denominator on the left side: $\frac{(c+a) + (b+c)}{(b+c)(c+a)} = \frac{3}{a+b+c}$ $\frac{a+b+2c}{bc + ab + c^2 + ac} = \frac{3}{a+b+c}$ Cross-multiplying: $(a+b+2c)(a+b+c) = 3(bc + ab + c^2 + ac)$ $(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$ $a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3bc + 3ab + 3c^2 + 3ac$ Simplifying by subtracting common terms from both sides: $a^2 + b^2 - c^2 = ab$ Using the Cosine Rule: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$ Substituting $a^2 + b^2 - c^2 = ab$: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$ Since $\cos C = \frac{1}{2}$,we have $\angle C = 60^{\circ}$.

In $\triangle ABC$,if $\frac{1}{b+c} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\angle C$ is equal to: (in $^{\circ}$)

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