37. In $\angle ABC$, $AB=6, BC=5, CA=4$. If $\kappa$ is on $BC$ and $\frac{B K}{K C}=\frac{3}{2}$, then $A K=(\quad$.
(A) $2 \sqrt{3}$
(B) 4
(C) $3 \sqrt{2}$
(D) $2 \sqrt{\frac{21}{5}}$
(E) *
37. (C).

Given $\frac{B K}{K C}=\frac{3}{2}, B C=5$. Therefore, $B D=3$.
$\because \cos B=\frac{A B^{2}+B C^{2}-A C^{2}}{2 \cdot A B \cdot B C}=\frac{A B^{2}+B K^{2}-A K^{2}}{2 \cdot A B \cdot B K}$.
$\therefore \frac{6^{2}+5^{2}-4^{2}}{2 \times 6 \times 5}=\frac{6^{2}+3^{2}-A K^{2}}{2 \times 6 \times 3}$. Hence, $A K=3 \sqrt{2}$.