①∵ AB切⊙O于点B
∴ OB ⊥ AB 则∠ABO = 90°
又∠BAO = 30° ,OA = 4
∴ OB = 2 ∠AOB = 60° AB = 2√3 ②∵ CB∥OA ∴ ∠CBO = ∠AOB = 60°
又OB = OC
∴△COB为正三角形 则∠COB = 60°
③ S阴影 = S△AOB +S扇形OCB - S△AOC
= 1/2 · 2 · 2√3 + 60π·2²/360 - 1/2 · 2 · 4 · sin120°
= 2√3 + 2π/3 - 2√3
= 2π/3 【sin120° = sin60°】