19. Given that the three sides of $\triangle A B C$ are all integers, $\cos A=\frac{11}{16}, \cos B=\frac{7}{8}, \cos C=-\frac{1}{4}$. Then the minimum possible value of the perimeter of $\triangle A B C$ is ( ).
(A) 9
(B) 12
(C) 23
(D) 27
(E) 44
19. A.

From the problem, we know
$$
\begin{array}{l}
\cos A=\frac{11}{16}, \sin A=\frac{3 \sqrt{15}}{16}, \\
\cos B=\frac{7}{8}, \sin B=\frac{\sqrt{15}}{8}, \\
\cos C=-\frac{1}{4}, \sin C=\frac{\sqrt{15}}{4} .
\end{array}
$$

By the Law of Sines, we get
$$
a: b: c=\sin A: \sin B: \sin C=3: 2: 4 \text {. }
$$

Therefore, the minimum possible perimeter of $\triangle A B C$ is
$$
3+2+4=9 \text {. }
$$