Given: ∠PCA = 110°, PC is tangent at C, AOB is diameter.
Step 1: OC ⊥ PC (radius ⊥ tangent)
∴ ∠OCA = 90°
Step 2: ∠PCO = ∠PCA − ∠OCA = 110° − 90° = 20°
∴ ∠BCO = ∠PCO = 20° ... (OC = OB = radius, so ∠OCB = ∠OBC; but since ∠PCO = 20°, and OC = OB, △OCB is isosceles)
Wait — ∠BCO = ∠OCA − ∠OCA... Let me use: ∠ACO = 90°, so ∠ACB = 90° (angle in semicircle).
∠BCO:
∠OCB = ∠OBC (OC = OB = radii, △OCB isosceles)
∠PCA = 110° ⟹ ∠ACO = 90° − ∠PCO...
Correct working:
∠OCP = 90° (radius ⊥ tangent)
∠OCA = ∠OCP − ∠PCA ... no, ∠PCA = 110° > 90°
So: ∠OCA = ∠PCA − ∠PCO
∠PCO = 90° ⟹ ∠OCA = 110° − 90° = 20°
Since OC = OA (radii), △OCA is isosceles ⟹ ∠OAC = ∠OCA = 20°
∠AOC = 180° − 20° − 20° = 140°
∠BOC = 180° − 140° = 40° (AOB is straight line)
Since OC = OB (radii), △OCB is isosceles:
∠BCO = ∠OBC = (180° − 40°)/2 = 70°
∠CBA:
∠CBA = ∠OBC = 70°
(Or: ∠ACB = 90°, angle in semicircle; ∠CAB = 20°; ∴ ∠CBA = 70°)
Answers: ∠BCO = 70°, ∠CBA = 70°
Source: Chapter 10, Sections 10.2 & 10.4
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