Question 141, Given in the tetrahedron D-ABC, $\angle \mathrm{ACB}=\angle \mathrm{ABD}=90^{\circ}, \mathrm{CA}=\mathrm{CB}, \angle \mathrm{BAD}=30^{\circ}$, if the projection of point $\mathrm{C}$ on the plane $\mathrm{ABD}$ is exactly on $\mathrm{AD}$, try to find the sine value of the dihedral angle $\mathrm{C}-\mathrm{AB}-\mathrm{D}$.
Question 141, Solution: As shown in the figure, let the projection of $\mathrm{C}$ on $\mathrm{AD}$ be $\mathrm{E}$, and draw $\mathrm{CF} \perp \mathrm{AB}$ at $\mathrm{F}$. Then $\mathrm{E}$ is on $\mathrm{AD}$, and $\angle \mathrm{CEA}=\angle \mathrm{CEF}=\angle \mathrm{CEB}=90^{\circ}$. Since $\mathrm{CB}^{2}=\mathrm{CE}^{2}+\mathrm{BE}^{2}=\mathrm{AC}^{2}-\mathrm{AE}^{2}+\mathrm{BE}^{2}$, it follows that $\mathrm{AE}=\mathrm{BE}$, so $E$ is the midpoint of $A D$.

Let $A C=1$, then $C F=\frac{\sqrt{2}}{2}, A E=\frac{\frac{\sqrt{2}}{2} \times 2}{\sqrt{3}}=\sqrt{\frac{2}{3}}$, hence $C E=\sqrt{\mathrm{AC}^{2}-\mathrm{AE}^{2}}=\sqrt{\frac{1}{3}}$. Therefore, the sine of the dihedral angle $C-A B-D$ is $\sin \angle C F E=\frac{C E}{C F}=\frac{2}{\sqrt{6}}=\frac{\sqrt{6}}{3}$.