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# 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°