In the given figure, PA and PB are tangents from external point P to a circle with centre C and Q is any point on the circle. Then the measure of $\angle AQB$ is
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Option A: 62½°
Since ∠APB = 55°, and AQBP is a cyclic quadrilateral (A, Q, B lie on circle; ∠CAP = ∠CBP = 90°), ∠ACB = 180° − 55° = 125°. The reflex ∠ACB = 360° − 125° = 235°. Since Q is on the major arc, ∠AQB = 235°/2 = 62½° (angle at circumference = half the central angle on the same arc... using the property that ∠AQB = ½ reflex ∠ACB... wait — ∠AQB = ½ × 125° = 62½°).
(Using: ∠ACB + ∠APB = 180°, so ∠ACB = 125°; angle at circumference = ½ central angle → ∠AQB = 125°/2 = 62½°.)
Answer: A — 62½°
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Explanation
- Since CA ⊥ PA and CB ⊥ PB (radius ⊥ tangent), in quadrilateral CAPB: ∠ACB = 360° − 90° − 90° − 55° = 125°.
- Q is on the major arc AB (opposite side to P). The inscribed angle theorem states ∠AQB = ½ × central angle ∠ACB = 125°/2 = 62½°.
- Key theorem to remember: angle subtended at centre = 2 × angle subtended at any point on the remaining arc.