Step 1: Understanding the Concept:
The problem provides a relation involving sides and cosines of angles of a triangle.
The standard approach is to convert all trigonometric ratios into expressions involving sides using the Cosine Rule.
Step 2: Key Formula or Approach:
Cosine Rule formulas:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
Substitute these into the given equation and simplify to find a relationship between the sides.
Step 3: Detailed Explanation:
Given equation:
\[ \frac{2\cos A}{a} + \frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc} + \frac{b}{ca} \]
Let's simplify the Right Hand Side (RHS) first:
\[ \text{RHS} = \frac{a \cdot a + b \cdot b}{abc} = \frac{a^2 + b^2}{abc} \]
Now, substitute the Cosine Rule into the Left Hand Side (LHS):
\[ \text{LHS} = \frac{2\left(\frac{b^2 + c^2 - a^2}{2bc}\right)}{a} + \frac{\left(\frac{a^2 + c^2 - b^2}{2ac}\right)}{b} + \frac{2\left(\frac{a^2 + b^2 - c^2}{2ab}\right)}{c} \]
\[ \text{LHS} = \frac{b^2 + c^2 - a^2}{abc} + \frac{a^2 + c^2 - b^2}{2abc} + \frac{a^2 + b^2 - c^2}{abc} \]
To add these fractions, take a common denominator of $2abc$:
\[ \text{LHS} = \frac{2(b^2 + c^2 - a^2) + (a^2 + c^2 - b^2) + 2(a^2 + b^2 - c^2)}{2abc} \]
Expand the numerator:
\[ \text{Numerator} = 2b^2 + 2c^2 - 2a^2 + a^2 + c^2 - b^2 + 2a^2 + 2b^2 - 2c^2 \]
Group like terms:
For $a^2$: $-2a^2 + a^2 + 2a^2 = a^2$
For $b^2$: $2b^2 - b^2 + 2b^2 = 3b^2$
For $c^2$: $2c^2 + c^2 - 2c^2 = c^2$
So, the numerator simplifies to $a^2 + 3b^2 + c^2$.
\[ \text{LHS} = \frac{a^2 + 3b^2 + c^2}{2abc} \]
Now, equate the simplified LHS and RHS:
\[ \frac{a^2 + 3b^2 + c^2}{2abc} = \frac{a^2 + b^2}{abc} \]
Multiply both sides by $2abc$ (since $a, b, c \neq 0$ for a valid triangle):
\[ a^2 + 3b^2 + c^2 = 2(a^2 + b^2) \]
\[ a^2 + 3b^2 + c^2 = 2a^2 + 2b^2 \]
Rearrange to solve for a relationship:
\[ 3b^2 - 2b^2 + c^2 = 2a^2 - a^2 \]
\[ b^2 + c^2 = a^2 \]
This equation $a^2 = b^2 + c^2$ represents the Pythagorean theorem for a triangle.
This means the triangle is right-angled, and the hypotenuse is side $a$.
Therefore, the angle opposite to side $a$, which is $\angle A$, must be $90^\circ$ or $\frac{\pi}{2}$ radians.
Step 4: Final Answer:
The angle $A$ is $\frac{\pi}{2}$.