In the given figure, ABCD is a rectangle whose diagonals intersect at O. Diagonal AC is produced to E and ∠ECD = 140°. Find the angles of △ OAB.

Given,

∠ECD = 140°.

ABCD is a rectangle.

⇒ ∠DCO + ∠DCE = 180°

⇒ ∠DCO = 180° - 140°

⇒ ∠DCO = 40°.

∠CAB = ∠DCA = 40° [Alternate angles are equal, as CD ∥ AB and AC is transversal]

From figure,

∠OAB = ∠CAB = 40°

OB = OA [∵ diagonals of a rectangle are equal and bisect each other]

∠OAB = ∠OBA = 40° [Angles opposite to equal sides in a triangle are equal.]

In △AOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 40° + 40° = 180°

⇒ ∠AOB = 180° - 80°

⇒ ∠AOB = 100°.

Hence, ∠OAB = 40°, ∠ABO = 40°, ∠AOB = 100°.