In $ riangle \mathrm{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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In $	riangle \mathrm{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

Aicharya · Accepted Answer

$\begin{aligned} & \frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c} \ & \frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}=\frac{3(a+b+c)}{a+b+c} \ & \frac{a}{b+c}+1+\frac{b}{c+a}+1=3 \ & \Rightarrow \frac{a}{b+c}+\frac{b}{c+a}=1 \ & \Rightarrow a(c+a)+b(b+c)=(b+c)(c+a) \ & \Rightarrow a c+a^2+b^2+b c=b c+a b+c^2+a c \ & \Rightarrow a^2+b^2-c^2=a b ...(i) \end{aligned}$ $	herefore \quad$ By cosine Rule, $\begin{aligned} & \cos C=\frac{a^2+b^2-c^2}{2 a b} \ & \cos C=\frac{a b}{2 a b} \ & \Rightarrow \cos C=\frac{1}{2} \ & \Rightarrow C=\frac{\pi}{3} \end{aligned}$ ...[From (i)]

In $\triangle \mathrm{ABC}$, with usual notations, if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$, then $m…

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