The two forces `2sqrt2N` and `xN` are acting at a point their resultant is perpendicular to `hatxN` and having magnitude of `sqrt6N`.The angle between the two forces and magnitude of `x` are.

Options

A. `theta=120^(@),X=sqrt2N`
B. `theta=30^(@),X=sqrt2N`
C. `theta=150^(@),X=sqrt3N`
D. `theta=150^(@),X=sqrt2N`

Explanation

To solve the problem, we need to find the angle between the two forces and the magnitude of \( x \). Let's break this down step by step.

### Step 1: Understand the Forces and Resultant
We have two forces:
- \( F_1 = 2\sqrt{2} \, \text{N} \)
- \( F_2 = x \, \text{N} \)

The resultant \( R \) of these two forces is given to be \( \sqrt{6} \, \text{N} \) and is perpendicular to the force \( F_2 \) (i.e., \( x \, \text{N} \)).

### Step 2: Use the Pythagorean Theorem
Since the resultant is perpendicular to \( F_2 \), we can use the Pythagorean theorem to relate the forces and the resultant:
\[ R^2 = F_1^2 + F_2^2 \]
Substituting the known values:
\[ (\sqrt{6})^2 = (2\sqrt{2})^2 + x^2 \]
\[ 6 = 8 + x^2 \]

### Step 3: Solve for \( x^2 \)
Rearranging the equation:
\[ x^2 = 6 - 8 \]
\[ x^2 = -2 \]
This indicates an error in our assumption or calculations. Let's check the angle between the forces.

### Step 4: Use the Angle Between Forces
Let \( \theta \) be the angle between the two forces. The resultant can also be expressed using the formula:
\[ R^2 = F_1^2 + F_2^2 + 2F_1F_2 \cos(\theta) \]
Substituting the known values:
\[ 6 = (2\sqrt{2})^2 + x^2 + 2(2\sqrt{2})(x) \cos(\theta) \]
This simplifies to:
\[ 6 = 8 + x^2 + 4\sqrt{2}x \cos(\theta) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ x^2 + 4\sqrt{2}x \cos(\theta) + 2 = 0 \]

### Step 6: Finding \( \theta \)
From the information that the resultant is perpendicular to \( F_2 \), we know that:
\[ \sin(\theta) = \frac{R}{F_1} = \frac{\sqrt{6}}{2\sqrt{2}} = \frac{\sqrt{3}}{2} \]
This implies:
\[ \theta = 60^\circ \text{ or } 120^\circ \]

### Step 7: Final Calculation for \( x \)
Using \( \theta = 120^\circ \) (since the resultant is perpendicular to \( F_2 \)), we can substitute back into the equation:
\[ x^2 + 4\sqrt{2}x \cos(120^\circ) + 2 = 0 \]
Since \( \cos(120^\circ) = -\frac{1}{2} \):
\[ x^2 - 2\sqrt{2}x + 2 = 0 \]

### Step 8: Solve the Quadratic Equation
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = -2\sqrt{2}, c = 2 \):
\[ x = \frac{2\sqrt{2} \pm \sqrt{(-2\sqrt{2})^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1} \]
\[ x = \frac{2\sqrt{2} \pm \sqrt{8 - 8}}{2} \]
\[ x = \frac{2\sqrt{2}}{2} = \sqrt{2} \]

### Final Answers
- The angle between the two forces is \( 120^\circ \).
- The magnitude of \( x \) is \( \sqrt{2} \, \text{N} \).