Question 2:
In the following figure, ∆ODC ∼ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB
Answer:
DOB is a straight line.
∴ ∠DOC + ∠COB = 180°
⇒ ∠DOC = 180° − 125°
= 55°
In ∆DOC,
∠DCO + ∠CDO + ∠DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that ∆ODC ∼ ∆OBA.
∴ ∠OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
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