Last lesson built the arrow from its shadows. This one runs the trade the other way: given the arrow's length $|\vec{v}|$ and its direction angle $\theta$ — degrees counterclockwise from the positive x-axis — the two shadows are $v_x = |\vec{v}|\cos\theta$ and $v_y = |\vec{v}|\sin\theta$. Read aloud: v-x equals the magnitude times cos theta, v-y equals the magnitude times sine theta. That is called resolving the vector, and $v_x$, $v_y$ come out in the same unit as $|\vec{v}|$ — the cosine and the sine are bare ratios and carry no unit of their own.

The swap is the most-marked error on the paper, so pin it with a case you cannot forget: at $\theta = 0^\circ$ the arrow lies flat along the x-axis, so $v_x$ must be the whole of it — and $\cos 0^\circ = 1$ while $\sin 0^\circ = 0$. Cosine owns the axis the angle is measured FROM. Two more rails: a shadow is never longer than the arrow, so any answer bigger than $|\vec{v}|$ is wrong before you check it; and past $90^\circ$ the cosine turns negative, which is not an error — it is the arrow pointing west, and the minus sign is the only thing that says so.