Question

In a triangle ABC, with usual notations if $\frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca}$ then $\angle A =$}

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

Hint:

Whenever an equation involves sides (a, b, c) and cosines, substituting the Cosine Rule usually reduces it to a simple algebraic form.

Answer 1

The Correct Option is A

Step 1: Concept
Use the cosine rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$, $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$, and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.

Step 2: Meaning
Substitute these into the given equation: $\frac{2(b^2+c^2-a^2)}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{2(a^2+b^2-c^2)}{2abc} = \frac{a^2+b^2}{abc}$.

Step 3: Analysis
Simplify the numerator: $(b^2+c^2-a^2) + \frac{a^2+c^2-b^2}{2} + (a^2+b^2-c^2) = a^2+b^2$. $b^2 + c^2 - a^2 + \frac{a^2}{2} + \frac{c^2}{2} - \frac{b^2}{2} + a^2 + b^2 - c^2 = a^2 + b^2$. $\frac{b^2}{2} + \frac{c^2}{2} + \frac{a^2}{2} = a^2 \implies b^2 + c^2 + a^2 = 2a^2 \implies b^2 + c^2 = a^2$.

Step 4: Conclusion
By Pythagoras theorem, the triangle is right-angled at A, so $\angle A = \pi/2$. Final Answer: (A)