$解:(2)过点A作AF⊥BC于点F，$
$则∠AFC=90°$
$因为∠BAC=67°，∠ABC=53°，$
$所以∠ACB=180°-∠BAC-∠ABC=60°.$
$因为AC=80m，$
$所以AF=AC·sin∠ACB=40\sqrt{3}m.$
$由题意,\frac {BC}{sin∠BAC}=\frac {AC}{sin∠ABC},$
$sin ∠BAC≈0.9，$
$得sin∠ABC≈0.8，$
$所以 BC≈90\ \mathrm {m}，$
$所以 S_{△ABC}=\frac {1}{2}×BC×AF=1800\sqrt{3}m².$
$故这片区域的面积约为1800\sqrt{3}m².$
