a: a⊥c
b⊥c
Do đó: a//b
b: Ta có: \(\hat{D_1}=\hat{D_2}\) (hai góc đối đỉnh)
mà \(\hat{D_1}=130^0\)
nên \(\hat{D_2}=130^0\)
Ta có: a//b
=>\(\hat{D_2}=\hat{C_1}\) (hai góc đồng vị)
=>\(\hat{C_1}=130^0\)
Ta có: \(\hat{C_1}+\hat{C_2}=180^0\) (hai góc kề bù)
=>\(\hat{C_2}=180^0-130^0=50^0\)
