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Question
In the following figure, ΔODC ∼ ΔOBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.
Solution
DOB is a straight line.
∴ ∠DOC + ∠COB = 180°
⇒ ∠DOC = 180° − 125°
⇒ ∠DOC = 55°
In ΔDOC,
∠DCO + ∠CDO + ∠DOC = 180°
(Sum of the measures of the angles of a triangle is 180°.)
⇒ ∠DCO + 70° + 55° = 180°
⇒ ∠DCO = 55°
It is given that ΔODC ∼ ΔOBA
∴ ∠OAB = ∠OCD ...[Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
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