From the given figure, DOB is a straight line.
$\angle DOC + \angle COB = 180°$
$\angle DOC$ = 180° – 125° (Given, $\angle BOC = 125°$) = 55°
Now in ΔDOC,
Sum of the measures of the angles of a triangle is 180º
$\angle DCO + \angle CDO + \angle DOC = 180°$
$\angle DCO + 70º + 55º = 180°$(Given, $\angle CDO = 70°$)
$\angle DCO = 180° - 125º = 55°$
It is given that, ΔODC ~ ΔOBA,
$\angle OAB = \angle OCD = 55°$
$\angle DOC = 55°, \angle DCO = 55°, \angle OAB = 55$