In the figure,$\Delta ODC \sim \Delta OBA$,$\angle BOC = 125^{\circ}$ and $\angle CDO = 70^{\circ}$. Find $\angle DOC$,$\angle DCO$,and $\angle OAB$.

(N/A) Since $DB$ is a straight line,$\angle DOC$ and $\angle BOC$ form a linear pair.
$\angle DOC + \angle BOC = 180^{\circ}$
$\angle DOC + 125^{\circ} = 180^{\circ}$
$\angle DOC = 180^{\circ} - 125^{\circ} = 55^{\circ}$
In $\triangle ODC$,the sum of the angles is $180^{\circ}$:
$\angle DCO + \angle CDO + \angle DOC = 180^{\circ}$
$\angle DCO + 70^{\circ} + 55^{\circ} = 180^{\circ}$
$\angle DCO + 125^{\circ} = 180^{\circ}$
$\angle DCO = 180^{\circ} - 125^{\circ} = 55^{\circ}$
Given that $\Delta ODC \sim \Delta OBA$,the corresponding angles are equal:
$\angle OAB = \angle OCD = \angle DCO$
Therefore,$\angle OAB = 55^{\circ}$.
Thus,$\angle DOC = 55^{\circ}$,$\angle DCO = 55^{\circ}$,and $\angle OAB = 55^{\circ}$.