Cube $ABCDEFGH$ is labeled so that $ABCD$ and $EFGH$ are opposite faces, with $AE$, $BF$, $CG$, and $DH$ as edges. Point $M$ is the midpoint of $\overline{AE}$, and point $N$ is the midpoint of $\overline{FG}$. What is $\cos(\angle MBN)$?

1. Let the cube have side length $2$ and coordinates $A=(0,0,0)$, $B=(2,0,0)$, $E=(0,0,2)$, $F=(2,0,2)$, and $G=(2,2,2)$. Then $M=(0,0,1)$ and $N=(2,1,2)$.
2. From $B$, $\overrightarrow{BM}=(-2,0,1)$ and $\overrightarrow{BN}=(0,1,2)$. Their dot product is $2$, and each has length $\sqrt5$.
3. Therefore $\cos(\angle MBN)=\dfrac{\overrightarrow{BM}\cdot\overrightarrow{BN}}{|\overrightarrow{BM}||\overrightarrow{BN}|}=\dfrac25$.