Find the magnitude and R, given forces F1=10, F2=15, F3=20, F4=50 and angles.

The x-component of a force $F$ at angle $\theta$ is $F_x = F \cos(\theta)$, and the y-component is $F_y = F \sin(\theta)$.

For $F_1 = 10$ at $\theta_1 = 30^\circ$:

$$F_{1x} = 10 \cos(30^\circ) = 10 \times \frac{\sqrt{3}}{2} \approx 8.660 \text{ units}$$ $$F_{1y} = 10 \sin(30^\circ) = 10 \times \frac{1}{2} = 5.000 \text{ units}$$

For $F_2 = 50$ at $\theta_2 = 270^\circ$:

$$F_{2x} = 50 \cos(270^\circ) = 50 \times 0 = 0 \text{ units}$$ $$F_{2y} = 50 \sin(270^\circ) = 50 \times (-1) = -50.000 \text{ units}$$

For $F_3 = 20$ at $\theta_3 = 240^\circ$:

$$F_{3x} = 20 \cos(240^\circ) = 20 \times (-\frac{1}{2}) = -10.000 \text{ units}$$ $$F_{3y} = 20 \sin(240^\circ) = 20 \times (-\frac{\sqrt{3}}{2}) \approx -17.321 \text{ units}$$

For $F_4 = 15$ at $\theta_4 = 150^\circ$:

$$F_{4x} = 15 \cos(150^\circ) = 15 \times (-\frac{\sqrt{3}}{2}) \approx -12.990 \text{ units}$$ $$F_{4y} = 15 \sin(150^\circ) = 15 \times \frac{1}{2} = 7.500 \text{ units}$$

$$\sum F_x = F_{1x} + F_{2x} + F_{3x} + F_{4x}$$ $$\sum F_x = 8.660 + 0 - 10.000 - 12.990 = -14.330 \text{ units}$$ $$\sum F_y = F_{1y} + F_{2y} + F_{3y} + F_{4y}$$ $$\sum F_y = 5.000 - 50.000 - 17.321 + 7.500 = -54.821 \text{ units}$$

$$R = \sqrt{(\sum F_x)^2 + (\sum F_y)^2}$$ $$R = \sqrt{(-14.330)^2 + (-54.821)^2}$$ $$R = \sqrt{205.3489 + 3005.349641}$$ $$R = \sqrt{3210.698541} \approx 56.663 \text{ units}$$

Since both $\sum F_x$ and $\sum F_y$ are negative, the resultant force lies in the third quadrant.
The reference angle $\alpha$ with the negative x-axis is given by:

$$\tan(\alpha) = \left| \frac{\sum F_y}{\sum F_x} \right| = \left| \frac{-54.821}{-14.330} \right| \approx 3.8256$$ $$\alpha = \arctan(3.8256) \approx 75.34^\circ$$

The angle $\theta_R$ measured counter-clockwise from the positive x-axis is:

$$\theta_R = 180^\circ + \alpha = 180^\circ + 75.34^\circ = 255.34^\circ$$

The magnitude of the resultant force is $\boxed{56.66 \text{ units}}$ and its direction is $\boxed{255.3^\circ}$ from the positive x-axis.