Question

如图在平面上将三角形ABC绕B点旋转到三角形A'BC'的位置时。AA’平行BC，角ABC=70°则角CBC'等于？

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Answer 1

解：∵AA'∥BC
∴∠BAA'=∠ABC=70°
∵AB=A'B
∴∠BA'A=∠BAA'=70°
∴∠ABA'=180°-70°-70°=40°
∵∠A'BC=∠ABC
∴∠A'BC-∠ABC'=∠ABC-∠ABC'
即∠CBC'=∠ABA'=40°

Answer 2

∵∠ABC=70°，AA‘∥BC
∴∠A'AB＝∠ABC=70°
∵BA=BA'
∴∠BA’A=∠A‘AB=70°
∴∠A’BA=40°
∴∠C’BC=∠A‘BA=40°