Question

ΔODC ~ΔOBA, $\angle$BOC = 125° and $\angle$CDO = 70°. Find $\angle$DOC, $\angle$DCO and $\angle$OAB.

Answer 1

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°