In $\triangle A B C$, if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$, then $C$ is equal to
In $\triangle A B C$, if $\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}$, then $C$ is equal to
- $90^{\circ}$
- $60^{\circ}$
- $45^{\circ}$
- $30^{\circ}$
Solution
Given that,
$
\begin{array}{ll}
& \frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c} \\
\Rightarrow & 1+\frac{b}{a+c}+1+\frac{a}{b+c}=3 \\
\Rightarrow & b(b+c)+a(a+c)=(a+c)(b+c) \\
\Rightarrow & b^2+b c+a^2+a c=a b+a c+b c+c^2 \\
\Rightarrow & a^2+b^2-c^2=a b \\
\text { Ne know that, } \cos C=\frac{a^2+b^2-c^2}{2 a b}=\frac{a b}{2 a b}=\frac{1}{2} \\
\Rightarrow \quad & C=60^{\circ}
\end{array}
$
Asked in: AP EAMCET 2008
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