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Solution: (i)

y = 100° {opposite angles of a parallelogram}

x + 100° = 180° {Adjacent angles of a parallelogram}

⇒ x = 180° – 100°

⇒ x = 80°

Solution (ii)

x = 90° {vertical opposite angles}

x + y + 30° = 180° {angle sum property of a triangle}

⇒ 90° + y + 30° = 180°

⇒ y = 180° – 120° = 60°

also, y = z = 60° {alternate angles}

Solution (¡¡¡)

z = 80° {corresponding angle}

z = y = 80° {alternate angles}

x + y = 180° {adjacent angles}

⇒ x + 80° = 180°

⇒ x = 180° – 80° = 100°

x + y + 30° = 180° {angle sum property of a triangle}

⇒ 90° + y + 30° = 180°

⇒ y = 180° – 120° = 60°

also, y = z = 60° {alternate angles}

x = z = 80° {opposite angles of a parallelogram}

Therefore, x = 80°, y = 100° and z = 80°

Solution (iv)

y = 112° {opposite angles of a parallelogram}

x = 180° – (y + 40°) {angle sum property of a triangle}

x = 28°

z = 28° {alternate angles}