In triangle ABC,with usual notations,if frac{ | vedclass

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(A) Given: $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=3$ $\frac{a}{b+c}+1+\f

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(A) Given: $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=3$ $\frac{a}{b+c}+1+\frac{b}{c+a}+1=3$ $\frac{a}{b+c}+\frac{b}{c+a}=1$ $a(c+a)+b(b+c)=(b+c)(c+a)$ $ac+a^2+b^2+bc=bc+ab+c^2+ac$ $a^2+b^2-c^2=ab$ By the Cosine Rule: $\cos C = \frac{a^2+b^2-c^2}{2ab}$ Substitute $a^2+b^2-c^2=ab$: $\cos C = \frac{ab}{2ab} = \frac{1}{2}$ Therefore,$C = \frac{\pi}{3}$.

In $\triangle ABC$,with usual notations,if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$,then $m \angle C$ is equal to

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