To determine the resultant force and its direction from the positive x-axis, we will resolve each force into its x and y components and then find the total force components in each direction.

Given:

1. $F_1 = 300 \, \text{N}$ at $30^\circ$ above the positive x-axis.
2. $F_2 = 600 \, \text{N}$ along the positive y-axis.
3. $F_3 = 450 \, \text{N}$ at $12^\circ$ below the negative x-axis.
4. $F_4 = 250 \, \text{N}$ at $60^\circ$ below the negative y-axis.

Step 1: Resolve each force into x and y components.

Force $F_1$:

• $F_{1x} = F_1 \cos(30^\circ) = 300 \cos(30^\circ)$
• $F_{1y} = F_1 \sin(30^\circ) = 300 \sin(30^\circ)$

Force $F_2$:

• $F_{2x} = 0$ (no horizontal component)
• $F_{2y} = 600 \, \text{N}$

Force $F_3$:

• $F_{3x} = -F_3 \cos(12^\circ) = -450 \cos(12^\circ)$
• $F_{3y} = -F_3 \sin(12^\circ) = -450 \sin(12^\circ)$

Force $F_4$:

• $F_{4x} = -F_4 \sin(60^\circ) = -250 \sin(60^\circ)$
• $F_{4y} = -F_4 \cos(60^\circ) = -250 \cos(60^\circ)$

Step 2: Calculate the total x and y components.

$F_{Rx} = F_{1x} + F_{2x} + F_{3x} + F_{4x}$ $F_{Ry} = F_{1y} + F_{2y} + F_{3y} + F_{4y}$

Step 3: Determine the magnitude of the resultant force.

$F_R = \sqrt{F_{Rx}^2 + F_{Ry}^2}$

Step 4: Find the direction of the resultant force.

The angle $\theta$ is given by: $\theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right)$

Let's proceed with calculating these values.