If tangents PA and PB from a point P to a circle with centre O are inclined to each other at angle of 80°, then POA is equal to (A) 50° (B) 60° (C) 70° (D) 80° Fig. 10.11 152
Step-by-Step Solution
Key Concept: Use the properties of tangents: (i) Tangents drawn from an external point are equal (PA = PB). (ii) A radius drawn to the point of tangency is perpendicular to the tangent (OA ⟂ PA and OB ⟂ PB). (iii) The angle between two tangents equals 180° minus the central angle subtended by the points of contact (∠APB = 180° – ∠AOB). By congruence of triangles OPA and OPB, the required angle POA is half of the central angle AOB.
1. Given: PA and PB are tangents from external point P to the circle with centre O. \(\angle APB = 80^{\circ}\).
2. Properties of tangents:
- \(PA = PB\) (tangents from a common external point are equal).
- \(OA \perp PA\) and \(OB \perp PB\) (radius ⟂ tangent at the point of contact).
3. Angle between tangents:
\[\angle APB = 180^{\circ} - \angle AOB\]
Hence \[\angle AOB = 180^{\circ} - 80^{\circ} = 100^{\circ}.\]
4. Congruence of triangles:
In triangles \(\triangle OPA\) and \(\triangle OPB\):
- \(OA = OB\) (radii of the same circle),
- \(PA = PB\) (tangents from the same point),
- \(OP\) is common.
Therefore, by the RHS (Side‑Side‑Side) criterion, \(\triangle OPA \cong \triangle OPB\).
5. From the congruence, corresponding angles are equal:
\[\angle POA = \angle POB.\]
Since \(\angle POA + \angle POB = \angle AOB = 100^{\circ}\), each of them is half of \(100^{\circ}\):
\[\angle POA = \frac{100^{\circ}}{2} = 50^{\circ}.\]
6. Hence, \(\angle POA = 50^{\circ}\).
7. Answer choice: (A) 50°.
Correct Answer: A