In a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.
Answer: x = 30, y = 40: ∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°.
Step-by-step solution
Idea: In a cyclic quadrilateral, opposite angles add up to 180°: ∠A + ∠C = 180° and ∠B + ∠D = 180°. This gives two linear equations in x and y.
- ∠A + ∠C = 180°: (x + 7) + (3y + 23) = 180 ⇒ x + 3y = 150 … (1).1 mark
- ∠B + ∠D = 180°: (y + 8) + (4x + 12) = 180 ⇒ 4x + y = 160 … (2).1 mark
- From (2): y = 160 − 4x. In (1): x + 480 − 12x = 150 ⇒ −11x = −330 ⇒ x = 30; y = 160 − 120 = 40.1 mark
- ∠A = 37°, ∠B = 48°, ∠C = 3(40) + 23 = 143°, ∠D = 4(30) + 12 = 132°.1 mark
∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°.
Check: 37 + 143 = 180 ✓, 48 + 132 = 180 ✓, and 37 + 48 + 143 + 132 = 360 ✓.
Answer to write in the exam
∠A + ∠C = 180° (opposite angles of a cyclic quadrilateral) ⇒ x + 3y = 150 … (1)
∠B + ∠D = 180° ⇒ 4x + y = 160 … (2)
From (2): y = 160 − 4x; in (1): −11x = −330 ⇒ x = 30, y = 40
∴ ∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°
Common mistakes that cost marks
- Adding adjacent angles (∠A + ∠B = 180°), which is not a property of cyclic quadrilaterals.
- Only using the angle sum 360°, which gives one equation, not two.
How this can come in the exam
Short answer (3 marks)
In a cyclic quadrilateral PQRS, ∠P = (2a + 4)°, ∠Q = (b + 3)°, ∠R = (2b + 10)°, ∠S = (4a − 5)°. Find a and b.
Show answer
∠P + ∠R = 180: 2a + 2b = 166 ⇒ a + b = 83 (1 mark). ∠Q + ∠S = 180: 4a + b = 182 (1 mark). Subtract: 3a = 99 ⇒ a = 33, b = 50 (1 mark).Try one yourself
In a cyclic quadrilateral, ∠A = x°, ∠C = y° and ∠A − ∠C = 40°. Find ∠A and ∠C.
Show answer
x + y = 180, x − y = 40 ⇒ ∠A = 110°, ∠C = 70°.
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