17. (11th "Hope Cup" Invitational Competition Question) In a regular triangular prism $A B C-A_{1} B_{1} C_{1}$, all edges are of equal length, and $D$ is the midpoint of $A A_{1}$. The angle formed by $B C_{1}$ and $C D$ is $\qquad$, and the dihedral angle between plane $B C D$ and plane $C D B_{1}$ is $\qquad$.
17. Fill in $90^{\circ}, \arcsin \frac{\sqrt{10}}{4}$. Reasoning: As shown in the right figure, let $B C_{1}$ and $B_{1} C$ intersect at $E$, then $E$ is the center of the square $B C C_{1} B_{1}$, and $E$ is the projection of point $D$ on the plane $B C C_{1} B_{1}$. Therefore, the projection of $D B_{1}$ on the plane $B C C_{1} B_{1}$ is $B_{1} E$, and since $B_{1} E \perp B C_{1}$, by the three perpendiculars theorem, we know that $B C_{1} \perp D B_{1}$. Thus, $B C_{1} \perp$ plane $D B_{1} C$, which means $B C_{1} \perp D C$, i.e., $B C_{1}$ and $C D$ form a $90^{\circ}$ angle.

Draw $E F \perp C D$ at $F$, and connect $F B$, then $F B \perp D C$, so $\angle E F B$ is the plane angle of the dihedral angle $B_{1}-D C-B$. Let the edge length be 2, then $B E=\sqrt{2}, D B=D C=\sqrt{5}, B F \cdot \sqrt{5}=2 \cdot 2$, thus $B F=\frac{4}{\sqrt{5}}, \sin \angle E F B=\frac{B E}{B F}=\sqrt{2} \cdot \frac{\sqrt{5}}{4}=\frac{\sqrt{10}}{4}$, hence $\angle E F B=\arcsin \frac{\sqrt{10}}{4}$.