To solve this problem, we'll break down the forces into their components, sum them up to find the resultant force, and then determine the magnitude and direction of the resultant. Let's go step by step.

Step 1: Break Down the Forces into Components

Each force vector can be broken down into its horizontal (x-axis) and vertical (y-axis) components.

1. $F_1 = 10 \, \text{N}, 45^\circ$ North of East:

   • $F_{1x} = 10 \times \cos(45^\circ)$
   • $F_{1y} = 10 \times \sin(45^\circ)$

2. $F_2 = 10 \, \text{N}, 10^\circ$ South of East:

   • $F_{2x} = 10 \times \cos(10^\circ)$
   • $F_{2y} = -10 \times \sin(10^\circ)$ (Negative because it's South)

3. $F_3 = 6 \, \text{N}, \text{East}$:

   • $F_{3x} = 6 \, \text{N}$
   • $F_{3y} = 0$ (No vertical component)

4. $F_4 = 6 \, \text{N}, 20^\circ$ South of West:

   • $F_{4x} = -6 \times \cos(20^\circ)$ (Negative because it's West)
   • $F_{4y} = -6 \times \sin(20^\circ)$ (Negative because it's South)

Step 2: Calculate the Components

1. $F_1$ Components:

   • $F_{1x} = 10 \times \cos(45^\circ) = 10 \times \frac{\sqrt{2}}{2} = 7.07 \, \text{N}$
   • $F_{1y} = 10 \times \sin(45^\circ) = 7.07 \, \text{N}$

2. $F_2$ Components:

   • $F_{2x} = 10 \times \cos(10^\circ) = 9.85 \, \text{N}$
   • $F_{2y} = -10 \times \sin(10^\circ) = -1.74 \, \text{N}$

3. $F_3$ Components:

   • $F_{3x} = 6 \, \text{N}$
   • $F_{3y} = 0 \, \text{N}$

4. $F_4$ Components:

   • $F_{4x} = -6 \times \cos(20^\circ) = -5.64 \, \text{N}$
   • $F_{4y} = -6 \times \sin(20^\circ) = -2.05 \, \text{N}$

Step 3: Sum the Components

Now, sum the x and y components separately to find the resultant components.

• Resultant $x$-component: $R_x = F_{1x} + F_{2x} + F_{3x} + F_{4x} = 7.07 + 9.85 + 6 - 5.64 = 17.28 \, \text{N}$

• Resultant $y$-component: $R_y = F_{1y} + F_{2y} + F_{3y} + F_{4y} = 7.07 - 1.74 + 0 - 2.05 = 3.28 \, \text{N}$

Step 4: Calculate the Magnitude of