In the figure, \( \triangle \) ODC \(\sim\) \( \triangle \) OBA, \( \angle \) BOC = 125° and \( \angle \) CDO = 70°. Find \( \angle \) DOC, \( \angle \) DCO and \( \angle \) OAB.

As we can see from the figure, DOB is a straight line.

Therefore, \( \angle \) DOC + \( \angle \) COB = 180°
\( \angle \) DOC = 180° – 125° (Given, \( \angle \) BOC = 125°)
= 55°

In \(\triangle\) DOC, Sum of the measures of the angles of a triangle is 180º

Therefore, \( \angle \) DCO + \( \angle \) CDO + \( \angle \) DOC = 180°
\( \angle \) DCO + 70º + 55º = 180°(Given, \( \angle \) CDO = 70°)
\( \angle \) DCO = 55°

It is given that, \(\triangle\) ODC \(\sim\) OBA,
Therefore, \(\triangle\) ODC ~ OBA.

Hence, Corresponding angles are equal in similar triangles

\( \angle \) OAB = \( \angle \) OCD
\( \angle \) OAB = 55°