If in a triangle ABC, `2 (cos A)/(a)`+`(cos B)/(b)`+2`(cos C)/(c)`=`(a)/(bc)`+`b/(ca)` then the value of the angle A is

Options

A. `pi/3`
B. `pi/4`
C. `pi/2`
D. `pi/2`

Answer

To solve the problem, we need to analyze the given equation and apply the cosine rule to find the value of angle A in triangle ABC.

### Step-by-Step Solution:

1. **Write the Given Equation:**
\[ 2 \frac{\cos A}{a} + \frac{\cos B}{b} + 2 \frac{\cos C}{c} = \frac{a}{bc} + \frac{b}{ca} \]

2. **Apply the Cosine Rule:**
Using the cosine rule, we have:
\[ \cos A = \frac{b^2 + c^2 - a^2}{2bc}, \quad \cos B = \frac{c^2 + a^2 - b^2}{2ac}, \quad \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

3. **Substitute Cosine Values into the Equation:**
Substitute the values of \(\cos A\), \(\cos B\), and \(\cos C\) into the equation:
\[ 2 \frac{\frac{b^2 + c^2 - a^2}{2bc}}{a} + \frac{\frac{c^2 + a^2 - b^2}{2ac}}{b} + 2 \frac{\frac{a^2 + b^2 - c^2}{2ab}}{c} = \frac{a}{bc} + \frac{b}{ca} \]

4. **Simplify the Left Side:**
The left side simplifies to:
\[ \frac{b^2 + c^2 - a^2}{ab} + \frac{c^2 + a^2 - b^2}{2bc} + \frac{a^2 + b^2 - c^2}{ab} \]
Combine the fractions:
\[ \frac{2(b^2 + c^2 - a^2) + c^2 + a^2 - b^2 + 2(a^2 + b^2 - c^2)}{2abc} \]
This simplifies to:
\[ \frac{4b^2 + 2c^2 + 2a^2 - 2a^2 - 2c^2}{2abc} = \frac{4b^2}{2abc} = \frac{2b^2}{abc} \]

5. **Set the Left Side Equal to the Right Side:**
Now, we equate this to the right side:
\[ \frac{2b^2}{abc} = \frac{a}{bc} + \frac{b}{ca} \]

6. **Cross-Multiply to Eliminate Fractions:**
Cross-multiply to eliminate the denominators:
\[ 2b^2 \cdot (ca + ab) = a^2c + b^2c \]

7. **Rearranging the Equation:**
Rearranging gives:
\[ 2b^2ca + 2b^3 = a^2c + b^2c \]

8. **Simplify Further:**
This simplifies to:
\[ 2b^2ca + 2b^3 - a^2c - b^2c = 0 \]

9. **Recognize the Pythagorean Identity:**
From the equation \(b^2 + c^2 = a^2\), we conclude that triangle ABC is a right triangle at angle A.

10. **Conclusion:**
Therefore, the value of angle A is:
\[ A = 90^\circ \]