In parallelogram ABCD, AP and AQ are perpendiculars from vertex of obtuse angle A as shown. If ∠x : ∠y = 2 : 1; find the angles of the parallelogram.

Given,

∠x : ∠y = 2 : 1

Let ∠x = 2a and ∠y = a.

From figure,

AQCP is a quadrilateral.

∴ ∠A + ∠P + ∠C + ∠Q = 360° (Sum of interior angles of a quadrilateral equals)

⇒ y + 90° + x + 90° = 360°

⇒ a + 90° + 2a + 90° = 360°

⇒ 3a + 180° = 360°

⇒ 3a = 360° - 180°

⇒ 3a = 180°

⇒ a = $\dfrac{180°}{3}$

⇒ a = 60°.

⇒ ∠x = 2 × 60° = 120° and ∠y = 60°.

From figure,

⇒ ∠C = ∠x = 120°,

⇒ ∠A = ∠C = 120° (Opposite angles of parallelogram are equal),

⇒ ∠B + ∠C = 180° (Sum of adjacent angles of a parallelogram equals to 180°)

⇒ ∠B + 120° = 180°

⇒ ∠B = 180° - 120° = 60°

⇒ ∠D = ∠B = 60° (Opposite angles of parallelogram are equal).

Hence, ∠DAB = ∠C = 120° and ∠B = ∠D = 60°.