4. In $\triangle A B C$, $\sin A: \sin B: \sin C=2: 3: 4$, then $\angle A B C=$ $\qquad$ (result expressed in terms of inverse trigonometric function values).
4. $\arccos \frac{11}{16}$ Hint: $a: b: c=\sin A: \sin B: \sin C=2: 3: 4$. Let's assume $a=2 k, b=3 k, c=4 k$, then $\cos \angle A B C=\frac{a^{2}+c^{2}-b^{2}}{2 a c}=\frac{(2 k)^{2}+(4 k)^{2}-(3 k)^{2}}{2 \times 2 k \times 4 k}=\frac{11}{16}$. Since $0<\angle A B C<\pi$, thus $\angle A B C=$ $\arccos \frac{11}{16}$.