6. As shown in Figure 1, with the side $AB$ of the acute triangle $\triangle ABC$ as the diameter, draw $\odot O$, intersecting $AC$ and $BC$ at points $E$ and $D$ respectively, and $DF \perp AB$. If
$$
\begin{array}{l}
AC=14, CD=4,7 \sin C \\
=3 \tan B, \text { then } DB= \\
(\quad) .
\end{array}
$$
(A) 3
(B) 4
(C) 5
(D)6
6.D.

As shown in Figure 1, connect $A D$. Since $A B$ is the diameter of $\odot O$, we have $\angle A D C=\angle A D B=90^{\circ}$.
In the right triangle $\triangle A C D$, by the Pythagorean theorem, we get
$$
A D=\sqrt{A C^{2}-C D^{2}}=\sqrt{180}=6 \sqrt{5} \text {. }
$$

Then $\sin C=\frac{A D}{A C}=\frac{3 \sqrt{5}}{7}$.
Therefore, in the right triangle $\triangle A B D$, we have $\tan B=\frac{A D}{D B}=\frac{7}{3} \times \frac{3 \sqrt{5}}{7}=\sqrt{5}$.
Thus, $D B=6$.