If in a triangle ABC,frac{1}{a+c} + frac{1}{b | vedclass

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

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$60$

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(C) Given the equation: $\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$ Taking $LCM$ on the left side: $\frac{(b+c) + (a+c)}{(a+c)(b+c)} = \frac{3}{a+b+c}$ $\Rightarrow \frac{a+b+2c}{ab + ac + bc + c^2} = \frac{3}{a+b+c}$ Cross-multiplying: $(a+b+2c)(a+b+c) = 3(ab + ac + bc + c^2)$ $(a+b)^2 + c(a+b) + 2c(a+b) + 2c^2 = 3ab + 3ac + 3bc + 3c^2$ $a^2 + b^2 + 2ab + 3ac + 3bc + 2c^2 = 3ab + 3ac + 3bc + 3c^2$ $a^2 + b^2 - ab = c^2$ Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C$ Comparing $a^2 + b^2 - ab = c^2$ with $a^2 + b^2 - 2ab \cos C = c^2$,we get: $ab = 2ab \cos C$ $\cos C = \frac{1}{2}$ Therefore,$\angle C = 60^{\circ}$.

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

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