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

如图,角1=角2=角3,且角BAC=70度,角DFE=50度,求角ABC的度数
2025-06-22 13:47:57
推荐回答(4个)
回答1:

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

回答2:

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

回答3:

你的图在哪里?

回答4:

没有图怎么解答呀?