If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

  • A
    $30$
  • B
    $45$
  • C
    $60$
  • D
    $90$

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