求∠ABC、∠BAC和∠CAF
题目:求∠ABC、∠BAC和∠CAF
解答
EF ∥ GH,两条直线平行
所以∠ABC = 65° [内错角]
∠CAE = ∠ACH = 100° (内错角)
∠BAC + 65° = ∠CAE = ∠ACH = 100°
所以∠BAC = 100° - 65° = 35°
∠CAF = 180° - 100° = 80° [同旁内角]
∠ABC = 65°
∠BAC = 35°
∠CAF = 80°
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题目:求∠ABC、∠BAC和∠CAF
解答
EF ∥ GH,两条直线平行
所以∠ABC = 65° [内错角]
∠CAE = ∠ACH = 100° (内错角)
∠BAC + 65° = ∠CAE = ∠ACH = 100°
所以∠BAC = 100° - 65° = 35°
∠CAF = 180° - 100° = 80° [同旁内角]
∠ABC = 65°
∠BAC = 35°
∠CAF = 80°