$解:過點(diǎn)?A?作?AH⊥BC?于?H?$
$∵?{S}_{△ABC}=27\ \mathrm {cm}2?$
$∴?\frac {1}{2}×9×AH=27?$
$∴?AH=6\ \mathrm {cm}?$
$∵?AB=10\ \mathrm {cm},??AH=6\ \mathrm {cm}?$
$∴?BH=\sqrt{AB2-AH2}=\sqrt{102-62}=8(\ \mathrm {cm})?$
$∴?tanB=\frac {AH}{BH}=\frac {6}{8}=\frac {3}{4}$