9. In a regular tetrahedron $S-ABC$, $M$ and $N$ are the midpoints of $SB$ and $SC$ respectively. If the section $AMN \perp$ side face $SBC$, then the dihedral angle between the side face and the base of the tetrahedron is $\qquad$
9. $\arccos \frac{\sqrt{6}}{6}$. 

Since plane $A M N \perp$ plane $S B C$, take the midpoint $G$ of $M N$, and connect $A G$, then $A G \perp M N$. Take the midpoint $D$ of $B C$, and connect $S D, A D$, then $\angle S D A$ is the plane angle of the dihedral angle to be found. Let the side length of the base be $a$, then $S A=$ $A D=\frac{\sqrt{3}}{2} a$. Therefore, $S D=\frac{\sqrt{2}}{2} a, G D=\frac{\sqrt{2}}{4} a$.
Thus, $\cos \angle S D A=\frac{\frac{\sqrt{2}}{4} a}{\frac{\sqrt{3}}{2} a}=\frac{\sqrt{6}}{6}$.