Let C be a circle with centre O and P be an external point to C. Let PA and PB be two tangents to C with A and B being the points of tangency, respectively. If PA and PB are inclined to each other at an angle of 60°, then find ∠POA.
Question:
Let C be a circle with centre O and P be an external point to C. Let PA and PB be two tangents to C with A and B being the points of tangency, respectively. If PA and PB are inclined to each other at an angle of 60°, then find ∠POA.
Let C be a circle with centre O and P be an external point to C. Let PA and PB be two tangents to C with A and B being the points of tangency, respectively. If PA and PB are inclined to each other at an angle of 60°, then find ∠POA.
Options
Answer: (2) 60°
Explanation:
PA and PB are tangents from external point P, so ∠PAO = 90° and ∠PBO = 90°. In quadrilateral OAPB: 90° + 90° + ∠APB + ∠AOB = 360° ∠APB + ∠AOB = 180° Given ∠APB = 60° So ∠AOB = 120°. Since PA = PB and OA = OB, triangle AOB is symmetric. Therefore, OP bisects ∠AOB. ∠POA = ½ × 120° = 60°. Final Answer: 60°
Explanation:
PA and PB are tangents from external point P, so ∠PAO = 90° and ∠PBO = 90°. In quadrilateral OAPB: 90° + 90° + ∠APB + ∠AOB = 360° ∠APB + ∠AOB = 180° Given ∠APB = 60° So ∠AOB = 120°. Since PA = PB and OA = OB, triangle AOB is symmetric. Therefore, OP bisects ∠AOB. ∠POA = ½ × 120° = 60°. Final Answer: 60°
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