- A55°
- B110°
- C35°
- D70°
Step-by-Step Solution
In quadrilateral OAPB, ∠OAP = ∠OBP = 90°. So ∠APB = 360° − 90° − 90° − 110° = 70°. The angles at O and P are supplementary.
Answer: D) 70°
PA and PB are tangents from P to a circle with centre O. If ∠AOB = 110°, find ∠APB.
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