Question

如图所示,CE平分∠ACD,F为CA延长线上一点,FG‖CE交AB于点G,∠ACD=120°,∠AGF=40°,你能求出

Answer 1

∵CE平分∠ACD,∠ACD＝120

∴∠ACE＝∠DCE＝∠ACD/2＝120/2＝60

∵FG∥CE

∴∠F＝∠ACE＝60

∵∠AGF＝40

∴∠BAC＝∠F+∠AGF＝60+40＝100

∵∠ACD＝∠B+∠BAC

∴∠B＝∠ACD-∠BAC＝120-100＝20°