The magnitude of scalar product of two vectors is `8` and of vector product is `8sqrt(3)`. The angle between them is:

Options

A. `30^(@)`
B. `60^(@)`
C. `120^(@)`
D. `150^(@)`

Explanation

To find the angle between two vectors given their scalar product and vector product, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- The scalar product (dot product) of two vectors **A** and **B** is given by:
\[ A \cdot B = |A| |B| \cos \theta \]
- The vector product (cross product) of two vectors **A** and **B** is given by:
\[ A \times B = |A| |B| \sin \theta \]

2. **Set Up the Equations**:
- From the problem, we know:
\[ A \cdot B = 8 \quad \text{(Equation 1)} \]
\[ A \times B = 8\sqrt{3} \quad \text{(Equation 2)} \]

3. **Substitute into the Equations**:
- From Equation 1:
\[ |A| |B| \cos \theta = 8 \]
- From Equation 2:
\[ |A| |B| \sin \theta = 8\sqrt{3} \]

4. **Divide Equation 2 by Equation 1**:
- This gives us:
\[ \frac{|A| |B| \sin \theta}{|A| |B| \cos \theta} = \frac{8\sqrt{3}}{8} \]
- Simplifying this, we find:
\[ \tan \theta = \sqrt{3} \]

5. **Determine the Angle**:
- The angles for which \(\tan \theta = \sqrt{3}\) are:
\[ \theta = 60^\circ \quad \text{and} \quad \theta = 120^\circ \]

6. **Conclusion**:
- Therefore, the angle between the two vectors can be either \(60^\circ\) or \(120^\circ\).

### Final Answer:
The angle between the two vectors is \(60^\circ\) or \(120^\circ\).