Theory question 108
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 below, O is the centre of the circle ACDB. If ∠CAO = 26° and ∠AOB = 130°, calculate: (i) ∠OBC; (ii) ∠COB
Show the model answer
(a) Given: arc AB subtends ∠AOB at the centre O and ∠ACB at C on the remaining arc. To prove: ∠AOB = 2∠ACB.
Join CO and produce it to D. OA = OC (radii), so ∠OAC = ∠OCA = x; the exterior angle ∠AOD = 2x.
Similarly OB = OC gives ∠OCB = ∠OBC = y and ∠BOD = 2y.
∠AOB = 2x + 2y = 2(x + y) = 2∠ACB. (Diagram, given, to prove, construction and proof required.)
(b)(i) ∠ACB = ½ × 130° = 65° (angle at centre = twice angle at circumference)
OA = OC, so ∠ACO = ∠CAO = 26°; ∠BCO = 65° − 26° = 39°
OB = OC, so ∠OBC = ∠BCO = 39°
(ii) ∠COB = 180° − 39° − 39° = 102°
Equal radii make isosceles triangles, so their base angles are equal.