Ta có: \(\hat{A_2}+\hat{A_3}=180^0\) (hai góc kề bù)
=>\(\hat{A_3}=180^0-120^0=60^0\)
Ta có: \(\hat{A_2}=\hat{A_4}\) (hai góc đối đỉnh)
mà \(\hat{A_2}=120^0\)
nên \(\hat{A_4}=120^0\)
a//b
=>\(\hat{A_2}=\hat{B_2}\) (hai góc đồng vị)
=>\(\hat{B_2}=120^0\)
Ta có: \(\hat{B_2}+\hat{B_1}=180^0\) (hai góc kề bù)
=>\(\hat{B_1}=180^0-120^0=60^0\)