In Figure (2) given below, ABCD is a rectangle and diagonals intersect at O. AC is produced to E. If ∠ECD = 146°, find the angles of △AOB.

From figure,

∠OCD = 180° - 146° = 34° (As AE is a straight line).

The diagonals of a rectangle are equal and bisect each other.

∴ OC = OD

From figure,

In △OCD,

OC = OD

⇒ ∠ODC = ∠OCD = 34° (Angles opposite to equal sides are equal in isosceles triangle)

⇒ ∠ODC + ∠OCD + ∠DOC = 180°

⇒ 34° + 34° + ∠DOC = 180°

⇒ ∠DOC = 180° - 68° = 112°.

In △AOB,

⇒ ∠AOB = ∠DOC = 112° (Vertically opposite angles are equal).

⇒ ∠OAB = ∠OCD = 34°

⇒ ∠OBA = ∠ODC = 34°

Hence, ∠AOB = 112°, ∠OAB = 34° and ∠OBA = 34°.