8. Through the two vertices $A, B$ of $\triangle A B C$, draw perpendicular lines $A D, B E$ on the same side of plane $A B C$, forming an equilateral triangle $C D E$. Given $A D=1, B E=2, D E=3$. Find the dihedral angle between plane $C D E$ and plane $A B C$.
8. Let $\angle D C A=\theta_{1}, \angle E C B=\theta_{2}, \angle D C E=\theta$, and the dihedral angle to be found is $\varphi$. Given $\sin \theta_{1}=\frac{1}{3}, \sin \theta_{2}=\frac{2}{3}$, and $\theta=60^{\circ}$, by Corollary 1 of Property 9, we find $\sin ^{2} \varphi=\frac{4}{9}$. Therefore, the dihedral angle between plane $C D E$ and plane $A B C$ is $\arcsin \frac{2}{3}$.