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If in a $\triangle A B C, \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}$, then $\angle C$ is equal to
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Verified Answer
The correct answer is:
$60^{\circ}$
In $\triangle A B C$
$$
\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
$$
Let $\angle C=60^{\circ}$, then
$$
\cos C=\frac{\pi}{3}
$$
$$
\begin{aligned}
& \Rightarrow a^2+b^2-c^2=a b \\
& \Rightarrow b^2+b c+a^2+a c=a b+a c+b c+c^2 \\
& \Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)
\end{aligned}
$$
Divide by $(a+c)(b+c)$ and add 2 on both sides, we get
$$
\begin{aligned}
& 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\
\Rightarrow \quad & \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
\end{aligned}
$$
So, $\angle C$ should be $60^{\circ}$.
$$
\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
$$
Let $\angle C=60^{\circ}$, then
$$
\cos C=\frac{\pi}{3}
$$
$$
\begin{aligned}
& \Rightarrow a^2+b^2-c^2=a b \\
& \Rightarrow b^2+b c+a^2+a c=a b+a c+b c+c^2 \\
& \Rightarrow b(b+c)+a(a+c)=(a+c)(b+c)
\end{aligned}
$$
Divide by $(a+c)(b+c)$ and add 2 on both sides, we get
$$
\begin{aligned}
& 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\
\Rightarrow \quad & \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}
\end{aligned}
$$
So, $\angle C$ should be $60^{\circ}$.
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