MHT CET · Maths · Properties of Triangles
In a triangle \(A B C\), with usual notations.
\(\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}\)
Then \(\angle \mathrm{A}=\)
- A \(\frac{\pi}{4}\)
- B \(\frac{\pi}{6}\)
- C \(\frac{\pi}{2}\)
- D \(\frac{\pi}{3}\)
Answer & Solution
Correct Answer
(C) \(\frac{\pi}{2}\)
Step-by-step Solution
Detailed explanation
\(2(b^2+c^2-a^2) + (a^2+c^2-b^2) + 2(a^2+b^2-c^2) = 2a^2 + 2b^2\) \(2b^2+2c^2-2a^2 + a^2+c^2-b^2 + 2a^2+2b^2-2c^2 = 2a^2 + 2b^2\)
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