Two forces are applied to an object. One force is horizontal with a magnitude of 8 N, and the second force has a magnitude of 5 N and is directed at 45 degrees above the horizontal. What is the result

Let $\mathbf{F_1}$ be the horizontal force of 8 N.
Let $\mathbf{F_2}$ be the force of 5 N inclined at $45^\circ$ to the horizontal.

For $\mathbf{F_1}$:
Since it's a horizontal force, its components are:

$$F_{1x} = 8 \text{ N}$$ $$F_{1y} = 0 \text{ N}$$

For $\mathbf{F_2}$:
The force is 5 N at $45^\circ$ to the horizontal.

$$F_{2x} = 5 \cos(45^\circ) = 5 \times \frac{\sqrt{2}}{2} \approx 5 \times 0.7071 = 3.5355 \text{ N}$$ $$F_{2y} = 5 \sin(45^\circ) = 5 \times \frac{\sqrt{2}}{2} \approx 5 \times 0.7071 = 3.5355 \text{ N}$$

Let the resultant force be $\mathbf{F_R}$.
Total x-component ($F_x$):

$$F_x = F_{1x} + F_{2x} = 8 \text{ N} + 3.5355 \text{ N} = 11.5355 \text{ N}$$

Total y-component ($F_y$):

$$F_y = F_{1y} + F_{2y} = 0 \text{ N} + 3.5355 \text{ N} = 3.5355 \text{ N}$$

The magnitude of the resultant force, $|\mathbf{F_R}|$, is found using the Pythagorean theorem:

$$|\mathbf{F_R}| = \sqrt{F_x^2 + F_y^2}$$ $$|\mathbf{F_R}| = \sqrt{(11.5355 \text{ N})^2 + (3.5355 \text{ N})^2}$$ $$|\mathbf{F_R}| = \sqrt{133.067 + 12.500}$$ $$|\mathbf{F_R}| = \sqrt{145.567}$$ $$|\mathbf{F_R}| \approx 12.065 \text{ N}$$

The direction $\theta$ is found using the arctangent function, measured from the horizontal axis:

$$\theta = \arctan\left(\frac{F_y}{F_x}\right)$$ $$\theta = \arctan\left(\frac{3.5355 \text{ N}}{11.5355 \text{ N}}\right)$$ $$\theta = \arctan(0.3065)$$ $$\theta \approx 17.03^\circ$$

Since both $F_x$ and $F_y$ are positive, the direction is $17.03^\circ$ above the horizontal.

The magnitude of the resultant force is $\boxed{12.07 \text{ N}}$ and its direction is $\boxed{17.03^\circ \text{ above the horizontal}}$.