\(\triangle ABC\) has sides \(a, b, c\) such that \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\). Find \(\angle C\)

**Clarification:** \(a\) is the side opposite to vertex \(A\) and \(b\) is the side opposite to side vertex \(B\) and \(c\) is the side opposite to vertex \(C\) as shown in the image above.

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