摘要:∵ , ∴ ∠B = 30°. ∵ ∠AOC = 2 ∠B , ∴ ∠AOC = 60°.∵ ∠D = 30°, ∴ ∠OAD = 180°- ∠D - ∠AOD = 90°. ∴ AD是⊙O的切线. (2) 解:∵ OA = OC .∠AOC = 60°, ∴ △AOC是等边三角形 . ∴ OA = AC = 6 .

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