Question:
ΔODC ~ΔOBA, \(\angle\)BOC = 125° and \(\angle\)CDO = 70°. Find \(\angle\)DOC, \(\angle\)DCO and \(\angle\)OAB.
ΔODC ~ΔOBA, \(\angle\)BOC = 125° and \(\angle\)CDO = 70°. Find \(\angle\)DOC, \(\angle\)DCO and \(\angle\)OAB.
Updated On: Jan 13, 2026
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Solution and Explanation
DOB is a straight line.
∴ \(\angle\)DOC + \(\angle\)COB = 180°
⇒ \(\angle\)DOC = 180° − 125° = 55°
In ∆DOC,
\(\angle\)DCO + \(\angle\)CDO + \(\angle\)DOC = 180° (The sum of the measures of the angles of a triangle is 180°)
⇒ \(\angle\)DCO + 70º + 55º = 180°
⇒ \(\angle\)DCO = 55°
It is given that ∆ODC ∼ ∆OBA.
∴ \(\angle\)OAB = \(\angle\)OCD [Corresponding angles are equal in similar triangles]
⇒ \(\angle\)OAB = 55°
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