If, in a `/_\ ABC`,
`(2 cos A)/a + (cos B)/(b) + (2 cos C)/(c) = a/(bc) + b/(ca)`,
then: `/_A = .....`

Answer

To solve the given equation in triangle \( ABC \):

\[ \frac{2 \cos A}{a} + \frac{\cos B}{b} + \frac{2 \cos C}{c} = \frac{a}{bc} + \frac{b}{ca} \]

we will follow these steps:

### Step 1: Substitute the cosine values
We know the cosine values in terms of the sides of the triangle:
- \( \cos A = \frac{b^2 + c^2 - a^2}{2bc} \)
- \( \cos B = \frac{c^2 + a^2 - b^2}{2ac} \)
- \( \cos C = \frac{a^2 + b^2 - c^2}{2ab} \)

Substituting these into the equation gives:

\[ \frac{2 \left(\frac{b^2 + c^2 - a^2}{2bc}\right)}{a} + \frac{\left(\frac{c^2 + a^2 - b^2}{2ac}\right)}{b} + \frac{2 \left(\frac{a^2 + b^2 - c^2}{2ab}\right)}{c} = \frac{a}{bc} + \frac{b}{ca} \]

### Step 2: Simplify the left-hand side
This simplifies to:

\[ \frac{(b^2 + c^2 - a^2)}{bc} \cdot \frac{2}{a} + \frac{(c^2 + a^2 - b^2)}{2ac} \cdot \frac{1}{b} + \frac{(a^2 + b^2 - c^2)}{ab} \cdot \frac{2}{c} \]

### Step 3: Combine fractions
Finding a common denominator for the left-hand side, which is \( 2abc \):

\[ \frac{2(b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + 2(a^2 + b^2 - c^2)}{2abc} \]

### Step 4: Simplify the numerator
Now we simplify the numerator:

\[ 2b^2 + 2c^2 - 2a^2 + c^2 + a^2 - b^2 + 2a^2 + 2b^2 - 2c^2 \]

Combining like terms gives:

\[ (2b^2 - b^2 + 2b^2) + (2c^2 - 2c^2 + c^2) + (-2a^2 + a^2 + 2a^2) = 3b^2 + c^2 + a^2 \]

### Step 5: Set equal to right-hand side
Now we equate this to the right-hand side:

\[ \frac{3b^2 + c^2 + a^2}{2abc} = \frac{a}{bc} + \frac{b}{ca} \]

### Step 6: Clear denominators
Multiply through by \( 2abc \):

\[ 3b^2 + c^2 + a^2 = 2a^2 + 2b^2 \]

### Step 7: Rearrange terms
Rearranging gives:

\[ 3b^2 + c^2 + a^2 - 2a^2 - 2b^2 = 0 \]

This simplifies to:

\[ b^2 + c^2 - a^2 = 0 \]

### Step 8: Recognize the Pythagorean theorem
This implies:

\[ a^2 = b^2 + c^2 \]

### Conclusion
Since this follows the Pythagorean theorem, triangle \( ABC \) is a right triangle with the right angle at \( A \). Therefore:

\[ \angle A = 90^\circ \]