$解：∵? BC=5CD，??CD=1.6\ \mathrm {m}?$
$∴? BC=5×1.6=8(\mathrm {m})?$
$∴? BC?的長(zhǎng)為?8\ \mathrm {m}?$
$選擇條件①：由題意，得? \frac {AB}{BC}=\frac {DC}{CE}?$
$∴?\frac {AB}{8} =\frac {1.6}{1}?$
$∴? AB=12.8?$
$∴ 旗桿?AB?的高度為?12.8\ \mathrm {m}?$
$選擇條件②：如圖，過點(diǎn)?D?作?DF⊥AB，?垂足為?F?$
$則?DC=BF=1.6\ \mathrm {m}，??DF=BC=8\ \mathrm {m}?$
$在?Rt△ADF ?中，?∠ADF=54.46°?$
$∴? AF=DF· tan 54.46°≈8×1.4=11.2(\mathrm {m})?$
$∴? AB=AF+BF=11.2+1.6=12.8(\mathrm {m})?$
$∴ 旗桿?AB?的高度約為?12.8\ \mathrm {m}?$
