TA có: a//b
=>\(\hat{A_1}=\hat{B_2};\hat{A_2}=\hat{B_1}\) (các cặp góc so le trong)
mà \(\hat{A_2}-\hat{B_2}=30^0\)
nên \(\hat{B_1}-\hat{A_1}=30^0\)
mà \(\hat{B_1}+\hat{A_1}=180^0\) (hai góc kề bù)
nên \(\hat{B_1}=\frac{180^0+30^0}{2}=105^0;\hat{A_1}=105^0-30^0=75^0\)