Mathematics
In given figure, △ ODC ~ △ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Triangles
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Answer
From figure,
DOB is a straight line. So, ∠DOC and ∠BOC forms a linear pair.
⇒ ∠DOC + ∠BOC = 180°
⇒ ∠DOC + 125° = 180°
⇒ ∠DOC = 180° - 125° = 55°.
In △DOC,
⇒ ∠DOC + ∠CDO + ∠DCO = 180° [By angle sum property of triangle]
⇒ 55° + 70° + ∠DCO = 180°
⇒ ∠DCO = 180° - 125° = 55°.
From figure,
⇒ ∠OAB = ∠DCO = 55°. [Alternate angles are equal]
Hence, ∠DOC = 55°, ∠DCO = 55°, ∠OAB = 55°.
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