9. As shown in Figure 4, it is known that the power pole $A B$ stands vertically on the ground, and its shadow falls exactly on the slope surface $C D$ and the ground $B C$. If $C D$ makes a $45^{\circ}$ angle with the ground, $\angle A=60^{\circ}, C D$ $=4 \mathrm{~m}, B C=(4 \sqrt{6}-2 \sqrt{2}) \mathrm{m}$, then the length of the power pole $A B$ is $\qquad$ m. (accurate to $0.1 \mathrm{~m}$)
9.8.5.

As shown in Figure 8. Extend $A D$ to intersect the ground at point $E$, and draw $D F \perp C E$ at point $F$. Since $\angle D C F=45^{\circ}, \angle A=60^{\circ}$, and $C D=4$, we have:
$$
\begin{array}{c}
C F=D F=2 \sqrt{2}, \\
E F=D F \cdot \tan 60^{\circ}=2 \sqrt{6} .
\end{array}
$$

Since $\frac{A B}{B E}=\tan 30^{\circ}=\frac{\sqrt{3}}{3}$, we have:
$$
\begin{aligned}
A B & =B E \times \frac{\sqrt{3}}{3}=(2 \sqrt{6}+2 \sqrt{2}+4 \sqrt{6}-2 \sqrt{2}) \times \frac{\sqrt{3}}{3} \\
& =6 \sqrt{2} \approx 8.5(\mathrm{~m}) .
\end{aligned}
$$