Example 22 (2001 Guangdong College Entrance Examination Question) In the quadrilateral pyramid $S-ABCD$ with a right trapezoid base, $\angle ABC=$ $90^{\circ}, SA \perp$ plane $ABCD, SA=AB=BC=1, AD=\frac{1}{2}$, find the tangent of the dihedral angle formed by faces $SCD$ and $SBA$.
Solve: As shown in Figure 8-12, establish a coordinate system, $D\left(0, \frac{1}{2}, 0\right), C(-1,1,0)$, $S(0,0,1), \overrightarrow{C D}=\left(1,-\frac{1}{2}, 0\right), \overrightarrow{D S}=\left(0,-\frac{1}{2}, 1\right)$. Since the plane $S A B \perp y$-axis, the normal vector of plane $S A B$ is $\vec{m}=(0,1,0)$. Let the normal vector of plane $S D C$ be $\vec{n}=(x, y, z)$. From $\vec{n} \cdot \overrightarrow{D S}=0$ and $\vec{n} \cdot \overrightarrow{C D}=0$, we get $\vec{n}=(1,2,1), \cos \langle\vec{m}, \vec{n}\rangle=\frac{\vec{m} \cdot \vec{n}}{|\vec{m}| \cdot|\vec{n}|}=\frac{\sqrt{6}}{3}$.
Clearly, the plane angle of the dihedral angle we are looking for is an acute angle $\theta$,
then $\theta=\langle\vec{m}, \vec{n}\rangle$, so $\tan \theta=\frac{\sqrt{2}}{2}$.