Multiple choice

Point P lies outside a circle with centre O. Tangents PA and PB are drawn to meet the circle at A and B respectively. If (\angle APB = 80^\circ), then (\angle OAB) is equal to:

  1. $140^\circ$
  2. $40^\circ$
  3. $70^\circ$
  4. $35^\circ$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since PA and PB are equal tangents from point P, triangle APB is isosceles with angles at A and B equal. With angle APB = 80°, each base angle is (180-80)/2 = 50°. Since OA is perpendicular to PA (radius-tangent property), angle OAP = 90-50 = 40°. Then angle OAB = 90-40 = 40°.

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