If tangents PA and PB drawn from an external point P to the circle with centre O are inclined to each other at an angle of $80^\circ$ as shown in the given figure, then the measure of $\angle POA$ is:
Generated by claude-sonnet-4-6 · 2026-06-15 10:22 · grounding rag
Model Answer
(B) 50°
Since OA ⊥ PA (radius ⊥ tangent), ∠OAP = 90°. In △OAP, ∠POA = 180° – 90° – ∠APO = 180° – 90° – 40° = 50° (∠APO = 80°/2 = 40°, as OP bisects ∠APB).
Source: Chapter 10, Section 10.3 (Exercise 10.2, Q.3)
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Explanation
- The key property used: OP bisects ∠APB (since tangents from external point are equal, △OAP ≅ △OBP), so ∠APO = 40°.
- Radius is perpendicular to tangent: ∠OAP = 90°.
- Angle sum in △OAP gives ∠POA = 50°.
- The textbook lists this exact question as Exercise 10.2 Q.3 with answer (A) 50°. Note the options in the textbook match option (B) 50° here.