16. In a right triangular prism $A B C-A_{1} B_{1} C_{1}$, $\angle A C B=90^{\circ}, A C=2 B C, A_{1} B \perp B_{1} C$, find the sine of the angle formed by $B_{1} C$ and the lateral face $A_{1} A B B_{1}$.
16. Taking $C$ as the origin, $C A, C B, C C_{1}$ as the $x, y, z$ axes to establish a rectangular coordinate system, with $|B C|=2, \left|C C_{1}\right|=a$, then $A(4,0,0), A_{1}(4,0, a), B(0,2,0), B_{1}(0,2, a)$. Since $A_{1} B \perp B_{1} C$, we have $\overrightarrow{\mathrm{BA}_{1}} \cdot \overrightarrow{\mathrm{CB}_{1}}=(4,-2, a) \cdot(0,2, a)=a^{2}-4=0, a=2$.

Let $\vec{n}=(x, y, z)$ be a normal vector to the plane $A_{1} A B B_{1}$, then $\vec{n} \cdot \overrightarrow{A_{1} B_{1}}=-4 x+2 y=0$,
$\vec{n} \cdot \overrightarrow{B B_{1}}=2 z=0$, yielding $\vec{n}=(1,2,0)$.
Since $B_{1} C=(0,-2,-2)$,
the sine of the angle $\alpha$ between $B_{1} C$ and the plane $A_{1} A B B_{1}$ is
$$
\sin \alpha=\left|\vec{n} \cdot \overrightarrow{B_{1} C}\right| /\left(|\vec{n}|\left|\overrightarrow{B C_{1}}\right|\right)=\frac{\sqrt{10}}{5} .
$$