# In the figure, $$\triangle$$ ODC ~ $$\triangle$$ OBA, $$\angle$$ BOC = 125 and $$\angle$$ CDO = 70. Find the $$\angle$$ DOC, $$\angle$$ DCO and $$\angle$$ OAB.

## Solution :

Since BD is a line and OC is a ray on it.

$$\angle$$ DOC + $$\angle$$ BOC = 180

So,  $$\angle$$ DOC + 125 = 180

$$\angle$$ DOC = 55

In triangle CDO, we have :

$$\angle$$ CDO + $$\angle$$ DOC + $$\angle$$ DCO = 180

70 + 55 + $$\angle$$ DCO = 180

$$\angle$$ DCO = 55

Given that $$\triangle$$ ODC ~ $$\triangle$$ OBA

$$\angle$$ ODC = $$\angle$$ OBA, $$\angle$$ OCD = $$\angle$$ OAB

$$\angle$$ OBA = 70 and $$\angle$$ OAB = 55

Hence, $$\angle$$ DOC = 55, $$\angle$$ DCO = 55 and $$\angle$$ OAB = 55