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Question:

In Fig, △ODC ∼ △OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:

Since, ∠COD + ∠COB = 180° ∴ ∠COD = 180° - 125° = 55°
Since, ∠COD + ∠ODC + ∠DCO = 180° ∴ ∠DCO = 180° - 70° - 55° = 55°
Since, ∠DCO = ∠OAB = Alternate angles ∴ ∠OAB = 55°