Calculate the resultant of the forces shown in the diagram.

We use the following trigonometric values:
$\cos(40^\circ) \approx 0.7660$, $\sin(40^\circ) \approx 0.6428$
$\cos(30^\circ) \approx 0.8660$, $\sin(30^\circ) = 0.5000$
$\cos(60^\circ) = 0.5000$, $\sin(60^\circ) \approx 0.8660$

• Force 1 ($F_1 = 12 \text{ N}$): $40^\circ$ above the negative x-axis. $$F_{1x} = -12 \cos(40^\circ) = -12 \times 0.7660 = -9.1920 \text{ N}$$ $$F_{1y} = 12 \sin(40^\circ) = 12 \times 0.6428 = 7.7136 \text{ N}$$
• Force 2 ($F_2 = 10 \text{ N}$): $30^\circ$ above the positive x-axis. $$F_{2x} = 10 \cos(30^\circ) = 10 \times 0.8660 = 8.6600 \text{ N}$$ $$F_{2y} = 10 \sin(30^\circ) = 10 \times 0.5000 = 5.0000 \text{ N}$$
• Force 3 ($F_3 = 9 \text{ N}$): Vertically downwards. $$F_{3x} = 0 \text{ N}$$ $$F_{3y} = -9.0000 \text{ N}$$
• Force 4 ($F_4 = 15 \text{ N}$): $60^\circ$ below the positive x-axis. $$F_{4x} = 15 \cos(60^\circ) = 15 \times 0.5000 = 7.5000 \text{ N}$$ $$F_{4y} = -15 \sin(60^\circ) = -15 \times 0.8660 = -12.9900 \text{ N}$$

$$R_x = F_{1x} + F_{2x} + F_{3x} + F_{4x}$$ $$R_x = -9.1920 \text{ N} + 8.6600 \text{ N} + 0 \text{ N} + 7.5000 \text{ N}$$ $$R_x = 6.9680 \text{ N}$$ $$R_y = F_{1y} + F_{2y} + F_{3y} + F_{4y}$$ $$R_y = 7.7136 \text{ N} + 5.0000 \text{ N} - 9.0000 \text{ N} - 12.9900 \text{ N}$$ $$R_y = -9.2764 \text{ N}$$

$$R = \sqrt{R_x^2 + R_y^2}$$ $$R = \sqrt{(6.9680 \text{ N})^2 + (-9.2764 \text{ N})^2}$$ $$R = \sqrt{48.553024 \text{ N}^2 + 86.051696 \text{ N}^2}$$ $$R = \sqrt{134.60472 \text{ N}^2}$$ $$R \approx 11.602 \text{ N}$$

The direction is given by $\tan \theta = \frac{R_y}{R_x}$.

$$\tan \theta = \frac{-9.2764 \text{ N}}{6.9680 \text{ N}}$$ $$\tan \theta \approx -1.3312$$

Since $R_x$ is positive and $R_y$ is negative, the resultant force is in the fourth quadrant.

$$\theta = \arctan(-1.3312)$$ $$\theta \approx -53.09^\circ$$

This means the angle is $53.1^\circ$ below the positive x-axis.

The resultant force is $\boxed{11.6 \text{ N at } 53.1^\circ \text{ below the positive x-axis}}$.