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如图,在平面上将三角形ABC绕点B旋转到三角形A'BC'的位置时,AA'‖BC, ∠ABC=70° ,则∠CBC'=?
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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°
∴∠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°
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