12.2.27 * As shown in the figure, in tetrahedron $ABCD$, $AB \perp BC, BC \perp CD, CD \perp AB$
(1) Identify the side face that is perpendicular to face $BCD$, and prove it;
(2) If $AB=BC=1, CD=x$, and the dihedral angle $C-AD-B$ has a plane angle $\alpha, \sin \alpha=f(x)$, find the expression for $f(x)$ and the range of values for $\alpha$
(1) From $A B \perp B C$ and $A B \perp C D$, we know $A B \perp$ plane $B C D$, so plane $A B C \perp$ plane $B C D$, and plane $A B D \perp$ plane $B C D$.
(2) Draw $C E \mid B D$, with the foot of the perpendicular at $E$, then $C E \perp$ plane $A B D$. Draw $E F \perp A D$, with the foot of the perpendicular at $F$, and connect $C F$, then $C F \perp A D, \angle C F E$ is the plane angle of the dihedral angle $C-A D-B$, i.e., $\alpha=\angle C F E$.
In the right triangle $\triangle B C D$, by $B D \cdot C E=B C \cdot C D$, we have
$$
C E=\frac{B C \cdot C D}{B D}=\frac{x}{\sqrt{1+x^{2}}}, D E=\sqrt{C D^{2}-C E^{2}}=\sqrt{\frac{x^{4}}{1+x^{2}}}.
$$

In the right triangle $\triangle A B D$, $A D=\sqrt{A B^{2}+B D^{2}}=\sqrt{2+x^{2}}$. It is easy to see that $\triangle D F E \backsim \triangle D B A$, so $\frac{D E}{D A}=\frac{E F}{A B}, E F=\frac{D E \cdot A B}{D A}=\frac{x^{2}}{\sqrt{\left(1+x^{2}\right)\left(2+x^{2}\right)}}$. In the right triangle $\triangle C E F$, $C F=$ $\sqrt{C E^{2}+E F^{2}}=\frac{\sqrt{2} x}{\sqrt{2+x^{2}}}$. Therefore, $\sin \alpha=\frac{C E}{C F}=\frac{1}{\sqrt{2}} \sqrt{\frac{2+x^{2}}{1+x^{2}}} \cdot-\frac{1}{\sqrt{2}} \sqrt{1+\frac{1}{1+x^{2}}}(x \in$ $\left.\mathbf{R}^{+}\right)$. Thus, $f(x)=\frac{1}{\sqrt{2}} \sqrt{1+\frac{1}{1+x^{2}}}\left(x \in \mathbf{R}^{+}\right)$.

It is easy to find that the range of $f(x)$ is $\left(\frac{\sqrt{2}}{2}, 1\right)$, so $\frac{\sqrt{2}}{2}<\sin \alpha<1$. Since $\alpha \in\left[0, \frac{\pi}{2}\right]$, the range of $\alpha$ is $\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$.