a: m⊥a
n⊥a
Do đó: m//n
b:m//n
=>\(\hat{N_1}=\hat{M_1}\) (hai góc đồng vị)
=>\(\hat{M_1}=120^0\)
Ta có: \(\hat{M_1}+\hat{M_2}=180^0\) (hai góc kề bù)
=>\(\hat{M_2}=180^0-120^0=60^0\)
Ta có: \(\hat{M_1}=\hat{M_3}\) (hai góc đối đỉnh)
mà \(\hat{M_1}=120^0\)
nên \(\hat{M_3}=120^0\)
