Find the size of each lettered angle in the following figure.

In △ADC,

AD = CD (Given)

∠DAC = ∠DCA = 42° (Opposite angles of equal sides are equal).

⇒ ∠DAC + ∠DCA + ∠ADC = 180°

⇒ 42° + 42° + y = 180°

⇒ y + 84° = 180°

⇒ y = 96°

In △AOB and △COB,

AO = OC (Diagonals bisect each other)

AB = BC (Given)

∠AOB = ∠COB = 90° (Diagonals are perpendicular to each other)

Hence, △AOB ≅ △COB by RHS congruence rule.

By C.P.C.T. we get,

⇒ ∠ABO = ∠CBO = 26°

In △AOB,

⇒ ∠AOB + ∠ABO + ∠OAB = 180° (Sum of angles of triangle = 180°)

⇒ 90° + 26° + x = 180°

⇒ 116° + x = 180°

⇒ x = 180° - 116° = 64°

Hence, x = 64° and y = 96°.