2. In the regular quadrilateral pyramid $S-A B C D$, it is known that the sine value of the dihedral angle $A-S B-D$ is $\frac{\sqrt{6}}{3}$. Then the angle formed by the skew lines $S A$ and $B C$ is . $\qquad$
2. $60^{\circ}$.

As shown in Figure 1, let the intersection of $AC$ and $BD$ be $O$, and the projection of $O$ on $SB$ be $E$. Then $OE \perp SB$.
Also, $AC \perp$ plane $SBD$, so $AC \perp SB$.
Thus, $SB \perp$ plane $AEO \Rightarrow AE \perp SB$.
Therefore, $\angle AEO$ is the plane angle of the dihedral angle $A-SB-D$.
Hence, $\sin \angle AEO = \frac{AO}{AE} = \frac{\sqrt{6}}{3}$.
Let $AB = a$, $SA = b$. Then $AO = \frac{\sqrt{2}}{2} a$.
In $\triangle SAB$,
$S_{\triangle SAB} = \frac{1}{2} b AE = \frac{1}{2} a \sqrt{b^2 - \frac{a^2}{4}}$
$\Rightarrow AE = \frac{a \sqrt{b^2 - \left(\frac{a}{2}\right)^2}}{b}$.
From $\sin \angle AEO = \frac{\frac{\sqrt{2}}{2} a}{\frac{a \sqrt{b^2 - \left(\frac{a}{2}\right)^2}}{b}} = \frac{\sqrt{6}}{3}$
$\Rightarrow a = b$.

Thus, each side face of the pyramid is an equilateral triangle.
Therefore, the required angle is $60^{\circ}$.