Question

Answer 1

a: a⊥BA

b⊥BA

Do đó: a//b

b: Ta có: \(\hat{C_1}+\hat{C_2}=180^0\) (hai góc kề bù)

=>\(\hat{C_2}=180^0-47^0=133^0\)

Ta có: a//b

=>\(\hat{C_2}+\hat{D_2}=180^0\) (hai góc trong cùng phía)

=>\(\hat{D_2}=180^0-133^0=47^0\)

Ta có: \(\hat{D_2}+\hat{D_1}=180^0\) (hai góc kề bù)

=>\(\hat{D_1}=180^0-47^0=133^0\)

Ta có: \(\hat{D_1}=\hat{D_3}\) (hai góc đối đỉnh)

mà \(\hat{D_1}=133^0\)

nên \(\hat{D_3}=133^0\)