In Fig. 1, AB is diameter of a circle centered at O. BC is tangent to the circle at B. If OP bisects the chord AD and $\angle AOP = 60°$, then find $m\angle C$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Since OP bisects chord AD, OP ⊥ AD (perpendicular from centre bisects the chord).
So, ∠OPD = 90°.
In △AOP: ∠AOP = 60°, ∠APO = 90°
∴ ∠OAD = 180° − 90° − 60° = 30°
Since AB is diameter, ∠ADB = 90° (angle in semicircle).
In △BDC: ∠DBC = 90° (BC is tangent, OB ⊥ BC, so ∠DBC = 90°).
In △ABC: ∠DAB = 30°, ∠ABC = 90°
∴ ∠C = 180° − 90° − 30° = 60°
Source: Chapter 10, Sections 10.2–10.3
Explanation
- Key property used: The perpendicular from the centre to a chord bisects it — so OP ⊥ AD, giving ∠OPA = 90°.
- From △AOP, find ∠OAD (= ∠DAB) = 30° using angle sum.
- BC is tangent at B ⇒ OB ⊥ BC ⇒ ∠ABС = 90°.
- Finally, angle sum in △ABC gives ∠C = 60°.
- Examiners award marks for correctly identifying OP ⊥ AD, finding ∠DAB, using tangent-radius perpendicularity, and the final angle calculation.