Answer
Horizontal Component: $F_x \approx 39.4$ N
Vertical Component: $F_y \approx 30.8$ N
Work Step by Step
We can visualize the force as a vector $\textbf{F}$ drawn from the origin in the first quadrant, with the x-axis coinciding with the horizontal level path. We can then see $\textbf{F}$ as the hypotenuse of a right triangle with $\theta = 38^{\circ}$ adjacent to a horizontal leg $F_x$ along the x-axis (which is the x-component) and opposite a vertical leg $F_y$ parallel to the y-axis (which is the y-component). Since $|\textbf{F}| = 50$ N is the magnitude of $\textbf{F}$ and thus the length of the hypotenuse, we can use trigonometry to find the components:
$F_x = |\textbf{F}| \cos \theta = 50$ N $\cos 38^{\circ} \approx 39.4$ N
$F_y = |\textbf{F}| \sin \theta = 50$ N $\sin 38^{\circ} \approx 30.8$ N
