4. A tetrahedron with four right-angled triangular faces is called a
“tetrahedron with four right angles”. If in the tetrahedron $S-A B C$, $\angle S A B=\angle S A C=\angle S B C=90^{\circ}$, then the right angle of the other face is $\qquad$
4. $\angle A B C$.

From $\angle S A B=\angle S A C=90^{\circ}$, we know $S A \perp A B, S A \perp A C$.
Thus, $S A \perp B C$.
From $\angle S B C=90^{\circ}$, we know $S B \perp B C$.
Thus, $B C \perp$ plane $S A B \Rightarrow B C \perp A B$.
Therefore, $\angle A B C=90^{\circ}$.