$?解:①在Rt△ABC中,?$
$?因?yàn)锳C=12,BC=16?$
$?所以AB=\sqrt{AC2+BC2}=20?$
$?所以tanA=\frac {BC}{AC}=\frac {16}{12}=\frac {4}{3}?$
$?tanB=\frac {AC}{BC}=\frac {12}{16}=\frac {3}{4}?$
$?②在Rt△ABC中,?$
$?因?yàn)锽C=8 , AB=\sqrt{73}?$
$?所以AC=\sqrt{AB2-BC2}=3?$
$所以tanA=\frac {BC}{AC}=\frac {8}{3},tanB=\frac {AC}{BC}=\frac {3}{8}$