4. In the tetrahedron $S-A B C$, it is known that $A B=A C$, $S B=S C$. Then the angle formed by the line $S A$ and $B C$ is $\qquad$
4. $\frac{\pi}{2}$.

As shown in Figure 4, take the midpoint $M$ of side $BC$, and connect $AM$ and $SM$.

Since $AB = AC$, $SB = SC$, therefore,
$AM \perp BC$,
$SM \perp BC$.
Thus, $BC \perp$ plane $SAM$, which means $BC \perp SA$.
Therefore, the angle formed by line $SA$ and $BC$ is $\frac{\pi}{2}$.