Question

In a triangle ABC , with usual notations. \(\frac{2 \cos \mathrm{~A}}{\mathrm{a}}+\frac{\cos \mathrm{B}}{\mathrm{b}}+\frac{2 \cos \mathrm{C}}{\mathrm{c}}=\frac{\mathrm{a}}{\mathrm{bc}}+\frac{\mathrm{b}}{\mathrm{ca}}\). Then \(\angle \mathrm{A}=\)

• A. \(\frac{\pi}{4}\)
• B. \(\frac{\pi}{6}\)
• C. \(\frac{\pi}{2}\)
• D. \(\frac{\pi}{3}\)

Hint:

Multiply by abc and replace each cosine with the cosine rule.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Cosine rule: \(2bc\cos A=b^2+c^2-a^2\), \(2ca\cos B=c^2+a^2-b^2\), \(2ab\cos C=a^2+b^2-c^2\).

Step 2: Multiply by abc
The equation becomes \(2bc\cos A+ca\cos B+2ab\cos C=a^2+b^2\).

Step 3: Substitute
\((b^2+c^2-a^2)+\frac{c^2+a^2-b^2}2+(a^2+b^2-c^2)=a^2+b^2\).

Step 4: Simplify
The left side is \(2b^2+\frac{a^2+c^2-b^2}2\), so \(3b^2+a^2+c^2=2a^2+2b^2\), giving \(b^2+c^2=a^2\). Hence \(A=\frac\pi2\), option (C).

Final Answer:
Angle A is \(\frac\pi2\). \[ \boxed{\text{(C) }\dfrac{\pi}{2}} \]