$解：(2)过点 O 作 OF⊥AC于点 F，则∠OFA=90°，AF=\frac {1}{2}AC$
$∵∠AOB=90°，∠BOC = 30°，∴∠AOC = ∠AOB +∠BOC=120°$
$∵OA=OC，∴∠OAC=∠OCA=\frac {1}{2}(180°-∠AOC)=30°$
$∴OF=\frac {1}{2}OA$
$∵OA=2，∴OF=1，∴AF=\sqrt {OA²-OF²}= \sqrt {3}$
$∴AC=2AF=2 \sqrt {3}$
$∵CD=\frac {1}{2}AD，∴CD=\frac {1}{3}AC=\frac {2\sqrt {3}}{3}$
$∴S_{△ACD}=\frac {1}{2}CD · OF=\frac {\sqrt {3}}{3}$
$∵OB=OA=2，∴S_{扇形OBC}=\frac {30π×2^2}{360}=\frac {π}{3}$
$∴线段 BD、线段CD 和\widehat{BC}围成的图形的面积S=S_{扇形OBC}-S_{△OCD}=\frac {π-\sqrt 3}3$
