题目如图,PAPAPA,PBPBPB分别是⊙O⊙O⊙O的切线,AAA,BBB分别为切点,点EEE是⊙O⊙O⊙O上一点,且∠AEB=60°∠AEB=60°∠AEB=60°,则∠P∠P∠P为()(\quad)()A.120°120°120°B.60°60°60°C.30°30°30°D.45°45°45°知识点:切线长定理章节:图形的性质 / 圆