Two components in, one heading out: $\theta = \tan^{-1}\left(\dfrac{v_y}{v_x}\right)$, read aloud theta equals inverse tan of v-y over v-x. Rise over run — $v_y$ is the northward component and $v_x$ the eastward one, both in the same unit, and the ratio is bare, which is why $\theta$ comes back in degrees no matter what those units were. It is the right triangle's own $\tan\theta = \dfrac{o}{a}$ — opposite over adjacent — with the vector's two shadows playing the legs.

Here is the trap that costs whole marks. Your calculator's $\tan^{-1}$ key only ever returns an angle between $-90^\circ$ and $90^\circ$ — the first and fourth quadrants — because $\dfrac{4}{-3}$ and $\dfrac{-4}{3}$ are the same number and it cannot tell them apart. So when $v_x$ is negative, the arrow points west and the honest answer is $180^\circ$ minus what the key gave you. Sketch the arrow first, every time; the sketch knows the quadrant and the keypad does not. (The two-argument form, $\operatorname{atan2}(v_y,\, v_x)$, is the same fix written into software.) And check your mode: this room works in degrees.