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Answer 1

Qua B, kẻ tia BM nằm giữa hai tia BA và BC sao cho BM//Ax//Cy

BM//Ax

=>\(\hat{xAB}+\hat{ABM}=180^0\) (hai góc trong cùng phía)(1)

Ta có: BM//Cy

=>\(\hat{MBC}+\hat{BCy}=180^0\) (hai góc trong cùng phía)(2)

Từ (1),(2) suy ra \(\hat{xAB}+\hat{ABM}+\hat{MBC}+\hat{BCy}=180^0+180^0=360^0\)

=>\(\hat{xAB}+\hat{yCB}+\hat{ABC}=360^0\)

\(\hat{BCy}=\frac65\cdot120^0=144^0\)

\(\hat{ABC}=360^0-120^0-144^0=240^0-144^0=96^0\)