Geometry Essentials · Superchallenge problem

Four rays $OA$, $OB$, $OC$, $OD$ leave point $O$ in that order around the full turn, dividing it into $\angle AOB$, $\angle BOC$, $\angle COD$, and $\angle DOA$ (these four add to $360^\circ$). The bisector of $\angle AOB$ and the bisector of $\angle COD$ are opposite rays. If $m\angle AOB = 100^\circ$ and $\angle BOC$ is $30^\circ$ more than twice $\angle COD$, find $m\angle DOA$.