Question

如图，已知D是△ABC的BC边延长线上一点，DF⊥AB交AB于F，交AC于E，∠A=49°，∠D=54°，求∠ACB的度数

Answer 1

解：
∵DF⊥AB
∴∠BFD＝90
∴∠B+∠D＝90
∵∠D＝54
∴∠B＝36
∵∠A＝49
∴∠ACB＝180-∠A-∠B＝180-36-49＝95°

Answer 2

因为DF⊥AB
所以∠BFD=90°
∠B=180°-∠BFD-∠D
=180°-90°-54°
=36°
∠ACB=180°-∠B-∠A
=180°-36°-49°
=95°
所以∠ACB=95°

Answer 3

∵DF垂直AB
∴∠DFB=∠DFA=90°
∵∠A=35° ∠A+∠FEA+∠AFE=180°
∴∠FEA=180°-∠A-∠AFE
=180°-90°-35°
=55°
∵∠FEA和∠CED是对顶角
∴∠CED=∠FEA=55°
∵∠D=42°∠D+∠C+∠CED=180°
∴∠ACD=180°-∠CED-∠D
=180°-55°-42°
=85°
答：∠ACD为85°。