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