$解：過點(diǎn)?C?作?CE⊥AD?交?AD?的延長線于點(diǎn)?E?$
$∵點(diǎn)?D?是?BC?的中點(diǎn)$
$∴?BD=CD?$
$又?∠BAD=∠DEC=90°，??∠ADB=∠CDE?$
$∴?△ABD≌△ECD?$
$∴?AB=CE，??AD=DE，??∠B=∠DCE，??∠EAC=150°-90°=60°?$
$設(shè)?DE=AD=x，?則?EC=AE · tan 60°=2\sqrt 3x，??CD=\sqrt {DE^2+EC^2}=\sqrt {13}x?$
$∴?sinB=sin∠DCE=\frac {DE}{CD}=\frac {\sqrt {13}}{13}?$
