Question

Answer 1

Câu 10:

a: AD⊥ CD

BC⊥CD

Do đó: AD//BC

b: Ta có: AD//BC

=>\(\hat{A_2}+\hat{B_1}=180^0\) (hai góc trong cùng phía)

=>\(\hat{A_2}=180^0-123^0=57^0\)

Ta có: \(\hat{A_1}+\hat{A_2}=180^0\) (hai góc kề bù)

=>\(\hat{A_1}=180^0-57^0=123^0\)

Câu 9:

a: AD⊥ CD

BC⊥CD

Do đó: AD//BC

b: AD//BC

=>\(\hat{A_2}=\hat{B_1}\) (hai góc so le trong)

=>\(\hat{A_2}=70^0\)

Ta có: \(\hat{A_2}+\hat{A_1}=180^0\) (hai góc kề bù)

=>\(\hat{A_1}=180^0-70^0=110^0\)