Four, $\mathrm{AC}$ is in the plane $\mathrm{M}$ of the dihedral angle $\mathrm{M}-\mathrm{EF}-\mathrm{N}$, forming a $45^{\circ}$ angle with $\mathrm{EF}$, and $\mathrm{AC}$ forms a $30^{\circ}$ angle with the plane $N$. Find the degree measure of the dihedral angle $M-E F-N$.
Four, Solution: Let $\mathrm{AB}$ be on plane $\mathrm{N}$, connect $\mathrm{BC}$, then $\angle \mathrm{ACB}$
$$
=30^{\circ} \text{. Also, construct BD } \perp \text{ EF, }
$$

then AD is perpendicular to EF. (By the Three Perpendiculars Theorem)
$\therefore \angle \mathrm{ADB}$ is the plane angle of the dihedral angle,
$$
\sin \angle \mathrm{ADB}=\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC} \sin 30^{\circ}}{\mathrm{AC} \sin 45^{\circ}}=\frac{\sqrt{2}}{2},
$$

thus, the plane angle of the dihedral angle is $45^{\circ}$.