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.

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Accepted Answer

Let us analyze the problem step by step:

1. Let O be the center of the circle, and P be the external point. PA and PB are tangents from P to the circle, touching at A and B respectively.

2. The lengths of PA and PB are equal (tangents from an external point to a circle are equal).

3. The angle between the tangents PA and PB at P is given as 60°.

4. The line segments OA and OB are radii, and each is perpendicular to the tangent at the point of contact. So, OA ⟂ PA and OB ⟂ PB.

5. Consider quadrilateral OAPB:
   - OA = OB (radii)
   - PA = PB (tangents from P)
   - OA ⟂ PA, OB ⟂ PB

6. Let us join O to P. We need to find ∠POA (the angle at O between OP and OA).

7. At point P, angle APB = 60° (given).

8. At point O, angle AOB is the angle subtended by arc AB at the center.

9. The key property: The angle between the tangents (angle APB) = 180° - angle AOB.
   - This is because quadrilateral OAPB is cyclic (since OA ⟂ PA and OB ⟂ PB, and PA = PB), and the sum of opposite angles in a cyclic quadrilateral is 180°.
   - So, ∠APB + ∠AOB = 180°
   - ∠AOB = 180° - 60° = 120°

10. Now, in triangle OAP:
    - OA ⟂ PA, so angle OAP = 90°
    - OP is the line joining O and P
    - We need to find ∠POA

11. In triangle OAP:
    - OA = radius
    - PA = tangent
    - OP = distance from center to external point
    - Angle at A is 90° (since OA ⟂ PA)

12. In triangle OAP, using the property of tangents:
    - The angle between OP and OA is half the angle subtended by arc AB at the center (i.e., ∠AOB/2)
    - So, ∠POA = 1/2 × ∠AOB = 1/2 × 120° = 60°
    - But this is incorrect; let's check the geometry more carefully.

13. Let's use triangle OAP:
    - OA ⟂ PA, so angle OAP = 90°
    - Similarly, OB ⟂ PB, so angle OBP = 90°
    - The angle at P between PA and PB is 60°
    - The angle at O between OA and OB is 120°

14. Consider triangle OAP:
    - OA is radius, OP is the line from center to external point, PA is tangent
    - Angle OAP = 90°
    - Angle at P is what we want to find (angle POA)

15. In triangle OAP:
    - OA ⟂ PA (angle OAP = 90°)
    - Let angle POA = θ
    - The sum of angles in triangle OAP: θ + 90° + angle at P = 180°
    - So, angle at P = 180° - 90° - θ = 90° - θ

16. But angle between OP and OA is θ, and similarly for OB and OP.
    - The angle between OA and OB is 120°, and OP bisects this angle (since OP is the angle bisector between OA and OB)
    - So, angle POA = 1/2 × angle AOB = 1/2 × 120° = 60°

17. But the correct geometric property is:
    - The angle between the tangents = difference between 180° and the angle subtended by the chord AB at the center.
    - So, angle between tangents = 180° - angle AOB
    - Given: angle between tangents = 60°
    - So, angle AOB = 120°
    - The angle between OP and OA (i.e., angle POA) = 1/2 × angle AOB = 60°

18. But let's check with a diagram: At point P, the angle between the tangents is 60°, and the angle at O between OA and OB is 120°. The angle between OP and OA is half of angle AOB, i.e., 60°.

19. However, the correct answer is 60° (option 1). Let's check the calculation:
    - The angle between the tangents = θ = 60°
    - The angle subtended at the center by chord AB = 2 × angle between OP and OA
    - Let angle POA = x
    - So, angle AOB = 2x
    - The angle between the tangents = 180° - angle AOB = 180° - 2x
    - So, 60° = 180° - 2x
    - 2x = 180° - 60° = 120°
    - x = 60°

20. Therefore, the correct answer is 60°. Let's double-check:
    - The angle between the tangents = 180° - angle AOB
    - Let angle POA = x
    - angle AOB = 2x
    - 60° = 180° - 2x
    - 2x = 120°
    - x = 60°

21. So, the correct answer is 60° (option 2).

Final Answer: 60° (option 2).

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