2. In the triangular prism $A B C-A_{1} B_{1} C_{1}$, the base edge lengths and the side edge lengths are all equal, $\angle B A A_{1}=\angle C A A_{1}=60^{\circ}$. Then the angle formed by the skew lines $A B_{1}$ and $B C_{1}$ is $\qquad$
2. $\arccos \frac{\sqrt{6}}{6}$.

In the plane $B B_{1} C_{1} C$, translate $B C_{1}$ to $D B_{1}$, then $\angle A B_{1} D$ is the angle we need to find.
Assume the base edge length and the side edge length of the triangular prism are both 1.
In $\triangle A B_{1} D$,
$$
\begin{array}{l}
A B_{1}=\sqrt{3}, B_{1} D=\sqrt{2}, A D=\sqrt{3}, \\
\cos \angle A B_{1} D=\frac{A B_{1}^{2}+B_{1} D^{2}-A D^{2}}{2 A B_{1} \cdot B_{1} D} \\
=\frac{2+3-3}{2 \times \sqrt{3} \times \sqrt{2}}=\frac{\sqrt{6}}{6} .
\end{array}
$$

Thus, $\angle A B_{1} D=\arccos \frac{\sqrt{6}}{6}$.