If the area of a triangle of sides `vecA & vecB` is equal to `(AB)/(4)`, then find the acute angle between `vecA & vecB`.

Explanation

To solve the problem, we need to find the acute angle between the vectors \(\vec{A}\) and \(\vec{B}\) given that the area of the triangle formed by these vectors is equal to \(\frac{AB}{4}\).

### Step-by-Step Solution:

1. **Understand the formula for the area of a triangle formed by two vectors**:
The area \(A\) of a triangle formed by two vectors \(\vec{A}\) and \(\vec{B}\) can be expressed as:
\[ A = \frac{1}{2} |\vec{A} \times \vec{B}| \]
where \(|\vec{A} \times \vec{B}|\) is the magnitude of the cross product of the vectors.

2. **Set up the equation using the given area**:
We are given that the area of the triangle is equal to \(\frac{AB}{4}\). Therefore, we can write:
\[ \frac{1}{2} |\vec{A} \times \vec{B}| = \frac{AB}{4} \]

3. **Simplify the equation**:
Multiply both sides by 2 to eliminate the fraction:
\[ |\vec{A} \times \vec{B}| = \frac{AB}{2} \]

4. **Express the magnitude of the cross product**:
The magnitude of the cross product can also be expressed in terms of the magnitudes of the vectors and the sine of the angle \(\theta\) between them:
\[ |\vec{A} \times \vec{B}| = |\vec{A}| |\vec{B}| \sin \theta \]
Substituting this into our equation gives:
\[ |\vec{A}| |\vec{B}| \sin \theta = \frac{AB}{2} \]

5. **Cancel the magnitudes**:
Here, we denote \(A = |\vec{A}|\) and \(B = |\vec{B}|\). The equation simplifies to:
\[ AB \sin \theta = \frac{AB}{2} \]
Assuming \(AB \neq 0\), we can divide both sides by \(AB\):
\[ \sin \theta = \frac{1}{2} \]

6. **Find the angle \(\theta\)**:
The sine of an angle is equal to \(\frac{1}{2}\) at two angles in the range of \(0^\circ\) to \(180^\circ\):
\[ \theta = 30^\circ \quad \text{or} \quad \theta = 150^\circ \]

7. **Determine the acute angle**:
Since we are asked for the acute angle, we select:
\[ \theta = 30^\circ \]

### Final Answer:
The acute angle between the vectors \(\vec{A}\) and \(\vec{B}\) is \(30^\circ\).