如图,角1=角2=角3,且角BAC=70度,角DFE=50度,求角ABC的度数

题目

如图,角1=角2=角3,且角BAC=70度,角DFE=50度,求角ABC的度数

答案

∠DEF=∠EAC+∠3
=∠EAC+∠1
=∠BAC
=70°
∴可得∠ABC=∠2+∠ABD
=∠1+∠ABD
=∠EDF
=180-∠DEF-∠DFE
=60°