Ta có: a//b
\(\Rightarrow\widehat{A_1}=\widehat{ADC}=48^0\)(đồng vị)
Ta có: a//b
\(\Rightarrow\widehat{B_1}=\widehat{BCD}=60^0\)(so le trong)
Ta có: \(\widehat{C_1}+\widehat{BCD}=180^0\)(kề bù)
\(\Rightarrow\widehat{C_1}=180^0-\widehat{BCD}=180^0-60^0=120^0\)
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