Q69 / 160 · AP EAMCET 2013 · Mathematics

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

  • A
    $30^{\circ}$
  • B
    $45^{\circ}$
  • Correct answer
    C
    $60^{\circ}$
  • D
    $90^{\circ}$

Solution

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 $60^{\circ}$.

Hence, the correct answer is option: 'C'.

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