10. Given in the rectangular prism $A B C D-A_{1} B_{1} C_{1} D_{1}$, the side face $A_{1} A D D_{1}$ is a square, $M$ is the midpoint of edge $C D$, and the angle between $A M$ and $C D_{1}$ is $\theta$. If $\sin \theta=\frac{\sqrt{78}}{9}$, then the value of $\frac{A A_{1}}{A B}$ is ( ).
(A) $\sqrt{2}$
(B) $\frac{2 \sqrt{2}}{3}$
(C) $\frac{\sqrt{2}}{2}$
(D) $\frac{\sqrt{3}}{2}$
10. A.

As shown in Figure 5, establish a coordinate system. Let
$D A=D D_{1}=a, D C=b$, then
$$
\begin{array}{l}
D_{1}(0,0, a), C(0, b, 0), \\
D_{1} C=(0, b,-a) ; A(a, 0,
\end{array}
$$
$0), M\left(0, \frac{b}{2}, 0\right), A M=$
$\left(-a, \frac{b}{2}, 0\right)$. Then
$$
\begin{array}{l}
\cos \theta=\frac{\boldsymbol{A M} \cdot \boldsymbol{D}_{1} \boldsymbol{C}}{|\boldsymbol{A M}| \cdot\left|\boldsymbol{D}_{1} \boldsymbol{C}\right|}=\frac{\frac{b^{2}}{2}}{\sqrt{a^{2}+\frac{b^{2}}{4}} \cdot \sqrt{a^{2}+b^{2}}} \\
=\frac{b^{2}}{\sqrt{4 a^{2}+b^{2}} \cdot \sqrt{a^{2}+b^{2}}} .
\end{array}
$$

From the given information,
$$
1-\left(\frac{b^{2}}{\sqrt{4 a^{2}+b^{2}} \cdot \sqrt{a^{2}+b^{2}}}\right)^{2}=\left(\frac{\sqrt{78}}{9}\right)^{2} \text {. }
$$

Simplifying, we get $\frac{a}{b}=\sqrt{2}$, which means $\frac{A A_{1}}{A B}=\sqrt{2}$.