题目如图,四边形ABCDABCDABCD内接于⊙O\odot O⊙O,连接COCOCO并延长交⊙O\odot O⊙O于点EEE,连接DEDEDE.若∠A=105∘\angle A=105^{\circ}∠A=105∘,∠DEC=60∘\angle DEC=60^{\circ}∠DEC=60∘,则∠OCB=( )\angle OCB=\left(\ \ \right)∠OCB=( )A.35∘35^{\circ}35∘B.40∘40^{\circ}40∘C.45∘45^{\circ}45∘D.50∘50^{\circ}50∘知识点:切线的判定章节:图形的性质 / 圆