如图,△ABE和△ACD是△ABC分别沿着AB,AC边翻折180°形成的,若∠BAC=150°,则∠θ的度数是()
A.60°
B.50°
C.40°
D.30°
万明成回答:
∵∠BAC=150°
∴∠ABC+∠ACB=30°
∵∠EBA=∠ABC,∠DCA=∠ACB
∴∠EBA+∠ABC+∠DCA+∠ACB=2(∠ABC+∠ACB)=60°,即∠EBC+∠DCB=60°
∴θ=60°.
故选A.
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如图,△ABE和△ACD是△ABC分别沿着AB,AC边翻折180°形成的,若∠BAC=150°,则∠θ的度数是()
A.60°
B.50°
C.40°
D.30°
万明成回答:
∵∠BAC=150°
∴∠ABC+∠ACB=30°
∵∠EBA=∠ABC,∠DCA=∠ACB
∴∠EBA+∠ABC+∠DCA+∠ACB=2(∠ABC+∠ACB)=60°,即∠EBC+∠DCB=60°
∴θ=60°.
故选A.