Take home assignment. Forces of 430N, 100N, 150N, and 800N acting on a block.

We'll consider forces to the right as positive and forces to the left as negative.

$$F_{net,x} = 750 \, \text{N} - 430 \, \text{N}$$ $$F_{net,x} = 320 \, \text{N} \text{ (to the right)}$$

We'll consider forces downwards as negative.

$$F_{net,y} = -100 \, \text{N} - 800 \, \text{N} - 1000 \, \text{N}$$ $$F_{net,y} = -1900 \, \text{N} \text{ (downwards)}$$

The magnitude of the net force is found using the Pythagorean theorem:

$$F_{net} = \sqrt{F_{net,x}^2 + F_{net,y}^2}$$ $$F_{net} = \sqrt{(320 \, \text{N})^2 + (-1900 \, \text{N})^2}$$ $$F_{net} = \sqrt{102400 \, \text{N}^2 + 3610000 \, \text{N}^2}$$ $$F_{net} = \sqrt{3712400 \, \text{N}^2}$$ $$F_{net} \approx 1926.76 \, \text{N}$$

The angle $\theta$ with respect to the positive x-axis (horizontal to the right) is given by:

$$\theta = \arctan\left(\frac{F_{net,y}}{F_{net,x}}\right)$$ $$\theta = \arctan\left(\frac{-1900 \, \text{N}}{320 \, \text{N}}\right)$$ $$\theta = \arctan(-5.9375)$$ $$\theta \approx -80.44^\circ$$

This means the force is $80.44^\circ$ below the positive horizontal axis.

The magnitude of the net force is $\boxed{1926.76 \, \text{N}}$ at an angle of $\boxed{80.44^\circ \text{ below the positive horizontal axis}}$.