Question

Giải gấp giúp mình ạ

Answer 1

Ta có: a//b

=>\(\widehat{A_1}=\widehat{B_3}\)(hai góc so le trong)

mà \(\widehat{A_1}=120^0\)

nên \(\widehat{B_3}=120^0\)

Ta có: \(\widehat{B_3}+\widehat{B_4}=180^0\)(hai góc kề bù)

=>\(\widehat{B_4}+120^0=180^0\)

=>\(\widehat{B_4}=180^0-120^0=60^0\)

Ta có: a//b

=>\(\widehat{D_2}=\widehat{C_4}\)(hai góc so le trong)

mà \(\widehat{D_2}=100^0\)

nên \(\widehat{C_4}=100^0\)

Ta có: \(\widehat{C_2}=\widehat{C_4}\)(hai góc đối đỉnh)

mà \(\widehat{C_4}=100^0\)

nên \(\widehat{C_2}=100^0\)