Multiple choice

PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 128°, then ∠OAB is equal to:

  1. 38°

  2. 64°

  3. 72°

  4. 62°
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B Correct answer
Explanation

PA and PB are tangents from P, so PA = PB and ∠OPA = ∠OPB = 90°. In quadrilateral OAPB, ∠AOB = 360° - 90° - 90° - 128° = 52°. In isosceles triangle OAB (OA = OB as radii), ∠OAB = (180° - 52°)/2 = 64°. Alternatively: ∠APB = 128°, so ∠OAB = 90° - (180° - 128°)/2 = 64°.

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