AP EAMCET · Maths · Properties of Triangles
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 \(45^{\circ}\)
- C \(60^{\circ}\)
- D \(90^{\circ}\)
Answer & Solution
Correct Answer
(C) \(60^{\circ}\)
Step-by-step Solution
Detailed explanation
In \(\triangle A B C\) \[ \frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c} \] Let \(\angle C=60^{\circ}\), then \[ \cos C=\frac{\pi}{3} \]…
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