TS EAMCET · Maths · Properties of Triangles
In \(\triangle A B C\), if \(\frac{1}{b+c}+\frac{1}{c+a}=\frac{3}{a+b+c}\), then \(C\) is equal to
- A \(90^{\circ}\)
- B \(60^{\circ}\)
- C \(45^{\circ}\)
- D \(30^{\circ}\)
Answer & Solution
Correct Answer
(B) \(60^{\circ}\)
Step-by-step Solution
Detailed explanation
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