Theory question 109
Circle Theorems(a) Prove that the angle which an arc of a circle subtends at the centre is twice that which it subtends at any point on the remaining part of the circumference.
(b) In the diagram above, O is the centre of the circle, ∠OQR = 32° and ∠MPQ = 15°. Calculate (i) ∠QPR; (ii) ∠MQO.
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(a) Given a circle centre O, arc BC subtending ∠BOC at O and ∠BAC at A on the circumference. Join AO and produce to N.
OA = OB (radii), so ∠OAB = ∠OBA = x; exterior angle ∠BON = 2x. Similarly ∠CON = 2y where y = ∠OAC.
∠BOC = 2x + 2y = 2(x + y) = 2∠BAC.
(b)(i) OQ = OR, so ∠ORQ = 32° and ∠QOR = 180° − 64° = 116°; ∠QPR = ½ × 116° = 58°
(ii) ∠MOQ = 2 × ∠MPQ = 30°; triangle MOQ is isosceles, so ∠MQO = (180° − 30°)/2 = 75°
Use the angle at the centre twice: once for chord QR and once for chord MQ.