- A116°
- B58°
- C64°
- D148°
Step-by-Step Solution
OP bisects ∠APB, so ∠APB = 64°. In quadrilateral OAPB, ∠OAP = ∠OBP = 90°, so ∠AOB = .
Answer: A) 116°
PA and PB are tangents from P to a circle with centre O, and ∠OPA = 32°. Find ∠AOB.
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