In a cyclic quadrilateral sum of opposite angles is 180°.

Given, in a cyclic quadrilateral ABCD ,

∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°

⇒ 2x + 4 + 2y + 10 = 180 and y + 3 + 4x – 5 = 180

⇒ x + y = 83 --------- (1) and y + 4x = 182 ---------- (2)

Subtracting eq1 from eq2.

⇒ y + 4x – x – y = 182 – 83

⇒ 3x = 99

⇒ x = 33

Substituting in eq1.

⇒ y = 50

∠A = 2 × 33 + 4 = 70°

∠B = 50 + 3 = 53°

∠C = 2 × 50 + 10 = 110°

∠D = 4 × 33 – 5 = 127°