The two forces `2sqrt(2)N` and xN are acting at a point their resultant is perpendicular to `hat(x)N` and having magnitude of `vec(6) N`. The angle between the two forces and magnitude of x are

Options

A. `theta = 120^@ , x = sqrt(2)N`
B. `theta = 30^@ , x = sqrt(2)N`
C. `theta = 150^@ , x = sqrt(3)N`
D. `theta = 150^@, x = sqrt(2)N`

Answer

To solve the problem, we need to find the angle between the two forces and the magnitude of the force \( x \) given that the resultant of the two forces is perpendicular to one of the forces and has a magnitude of \( 6 \, \text{N} \).

### Step-by-step Solution:

1. **Understanding the Forces**:
We have two forces acting at a point:
- \( F_1 = 2\sqrt{2} \, \text{N} \)
- \( F_2 = x \, \text{N} \)

The resultant force \( R \) is given as \( 6 \, \text{N} \) and is perpendicular to \( F_2 \).

2. **Using the Pythagorean Theorem**:
Since the resultant \( R \) is perpendicular to \( F_2 \), we can use the Pythagorean theorem:
\[ R^2 = F_1^2 + F_2^2 \]
Substituting the known values:
\[ 6^2 = (2\sqrt{2})^2 + x^2 \]
\[ 36 = 8 + x^2 \]
\[ x^2 = 36 - 8 = 28 \]
\[ x = \sqrt{28} = 2\sqrt{7} \, \text{N} \]

3. **Finding the Angle Between the Forces**:
To find the angle \( \theta \) between the two forces, we can use the sine rule. Since the resultant is perpendicular to \( F_2 \), we can write:
\[ \sin(\theta) = \frac{R}{F_1} \]
\[ \sin(\theta) = \frac{6}{2\sqrt{2}} \]
Simplifying this:
\[ \sin(\theta) = \frac{6}{2\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]
However, we need to find the angle \( \theta \) such that:
\[ \sin(\theta) = \frac{\sqrt{3}}{2} \]
This corresponds to \( \theta = 60^\circ \) or \( 180^\circ - 60^\circ = 120^\circ \).

4. **Conclusion**:
Therefore, the angle between the two forces is \( 120^\circ \) and the magnitude of \( x \) is \( 2\sqrt{7} \, \text{N} \).

### Final Answer:
- Angle between the two forces: \( 120^\circ \)
- Magnitude of \( x \): \( 2\sqrt{7} \, \text{N} \)