If 2 cos (A-B) = 2 sin ( A+ B) `= sqrt3 ` find the value of acute angles A and B .

Answer

To solve the problem, we start with the equations given:

1. \( 2 \cos(A - B) = \sqrt{3} \)
2. \( 2 \sin(A + B) = \sqrt{3} \)

### Step 1: Simplify the equations
From the first equation:
\[ 2 \cos(A - B) = \sqrt{3} \]
Dividing both sides by 2 gives:
\[ \cos(A - B) = \frac{\sqrt{3}}{2} \]

From the second equation:
\[ 2 \sin(A + B) = \sqrt{3} \]
Dividing both sides by 2 gives:
\[ \sin(A + B) = \frac{\sqrt{3}}{2} \]

### Step 2: Determine angles from trigonometric values
Now we know:
\[ \cos(A - B) = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin(A + B) = \frac{\sqrt{3}}{2} \]

The angles corresponding to these values are:
- For \( \cos(A - B) = \frac{\sqrt{3}}{2} \), the angle \( A - B \) can be \( 30^\circ \) (since we are looking for acute angles).
- For \( \sin(A + B) = \frac{\sqrt{3}}{2} \), the angle \( A + B \) can be \( 60^\circ \).

### Step 3: Set up equations
Now we have two equations:
1. \( A - B = 30^\circ \) (Equation 1)
2. \( A + B = 60^\circ \) (Equation 2)

### Step 4: Solve the system of equations
To solve for \( A \) and \( B \), we can add the two equations:
\[ (A - B) + (A + B) = 30^\circ + 60^\circ \]
This simplifies to:
\[ 2A = 90^\circ \]
Dividing both sides by 2 gives:
\[ A = 45^\circ \]

Now substitute \( A = 45^\circ \) back into Equation 1:
\[ 45^\circ - B = 30^\circ \]
Solving for \( B \):
\[ B = 45^\circ - 30^\circ = 15^\circ \]

### Final Answer
Thus, the values of the acute angles are:
\[ A = 45^\circ, \quad B = 15^\circ \]