$解:延长AC,BD相交于点G，过点D作DF⊥AG于点F$
$∵斜坡CD的坡度i=1: \sqrt{3}，即 tan∠DCF=\frac{1}{3}=\frac{\sqrt{3}}{3}$
$∴∠DCF=30°$
$又∵CD=3.2m，∴DF=CDsin∠DCF=3.2×\frac{1}{2}=1.6m$
$CF=CDcos∠DCF=\frac {8}{5}\sqrt {3}m$
$由题意，易得\frac{DF}{FG}=\frac{1}{0.8}，∴FG=1.28m\ $
$∵AC=8.8m，∴AG=AC+CF+FG=(10.08+\frac {8}{5}\sqrt {3})m$
$∵易得\frac{AB}{AG}=\frac{1}{0.8}，∴AB=12.6+2\sqrt{3}≈16m$
$∴树高AB约为16m$
