In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.

Let ∠A = 3x and ∠C = x.

In cyclic quadrilateral ABCD,

⇒ ∠A + ∠C = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ 3x + x = 180°

⇒ 4x = 180°

⇒ x = $\dfrac{180°}{4}$ = 45°.

∠A = 3x = 3 x 45 = 135°

∠C = x = 45°.

Let ∠B = y and ∠D = 5y.

In cyclic quadrilateral ABCD,

⇒ ∠B + ∠D = 180° [Sum of opposite angles in a cyclic quadrilateral = 180°]

⇒ y + 5y = 180°

⇒ 6y = 180°

⇒ y = $\dfrac{180°}{6}$ = 30°.

∠B = y = 30°

∠D = 5y = 5 x 30° = 150°.

Hence, ∠A = 135°, ∠B = 30°, ∠C = 45° and ∠D = 150°.