MHT CET · Maths · Properties of Triangles
In \(\triangle A B C\), with usual notations, if \(\cos \frac{B}{2}=\sqrt{\frac{c+a}{2 a}}\), then \(a^2=\)
- A \(b^2-c^2\)
- B \(b+c\)
- C \(b^2+c^2\)
- D \(b-c\)
Answer & Solution
Correct Answer
(C) \(b^2+c^2\)
Step-by-step Solution
Detailed explanation
\(\cos \frac{B}{2}=\sqrt{\frac{s(s-b)}{ac}}\) \(\sqrt{\frac{s(s-b)}{ac}} = \sqrt{\frac{c+a}{2 a}}\)
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