Let a, b, c be sides of a triangle ABC and `Delta` denotes its area .
If `a=2, Delta=sqrt(3) and a cosC+sqrt(3) a sinC-b-c=0`, then find the value of `(b+c)`.
(symbols used have usual meaning in `DeltaABC`).

Answer

To solve the given problem step by step, we will use the information provided in the question.

### Step 1: Understand the given values
We have:
- \( a = 2 \)
- \( \Delta = \sqrt{3} \)
- The equation \( a \cos C + \sqrt{3} a \sin C - b - c = 0 \)

### Step 2: Use the area formula for a triangle
The area \( \Delta \) of triangle ABC can also be expressed using the formula:
\[ \Delta = \frac{1}{2} a b \sin C \]
Substituting the known values:
\[ \sqrt{3} = \frac{1}{2} \cdot 2 \cdot b \cdot \sin C \]
This simplifies to:
\[ \sqrt{3} = b \sin C \]
Thus, we can express \( b \) in terms of \( \sin C \):
\[ b = \frac{\sqrt{3}}{\sin C} \]

### Step 3: Substitute \( b \) in the given equation
Now, substitute \( b \) into the equation \( a \cos C + \sqrt{3} a \sin C - b - c = 0 \):
\[ 2 \cos C + \sqrt{3} \cdot 2 \sin C - \frac{\sqrt{3}}{\sin C} - c = 0 \]
This simplifies to:
\[ 2 \cos C + 2\sqrt{3} \sin C - c - \frac{\sqrt{3}}{\sin C} = 0 \]

### Step 4: Rearranging the equation
Rearranging gives us:
\[ c = 2 \cos C + 2\sqrt{3} \sin C - \frac{\sqrt{3}}{\sin C} \]

### Step 5: Find \( b + c \)
Now we need to find \( b + c \):
\[ b + c = \frac{\sqrt{3}}{\sin C} + \left(2 \cos C + 2\sqrt{3} \sin C - \frac{\sqrt{3}}{\sin C}\right) \]
This simplifies to:
\[ b + c = 2 \cos C + 2\sqrt{3} \sin C \]

### Step 6: Analyze the triangle
To find specific values, we can consider the case when the triangle is equilateral. In an equilateral triangle, \( a = b = c \) and all angles \( C = 60^\circ \).

### Step 7: Calculate \( b + c \) for equilateral triangle
If \( a = 2 \), then \( b = 2 \) and \( c = 2 \):
\[ b + c = 2 + 2 = 4 \]

### Conclusion
Thus, the value of \( b + c \) is:
\[ \boxed{4} \]