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