Question

 

Answer 1

3.19 

a) Ta có: 

\(\widehat{BMz}=\widehat{ANM}\left(=60^o\right)\)

Mà hai góc này ở vị trí đồng vị

=> Ax//By

b) Ax//By 

=> \(\widehat{ABy'}=\widehat{BAN}=50^o\) 

Vì hai góc này so le trong 

c) Ax//By 

\(=>\widehat{BAN}+\widehat{ABM}=180^o\)

Hai góc trong cùng phía

\(=\widehat{ABM}=180^o-\widehat{BAN}=180^o-50^o=130^o\)

Answer 2

3.20: b,d

 

3.19:

a: a//b

=>\(\widehat{A_1}=\widehat{B_1}\)(hai góc so le trong)

=>\(\widehat{A_1}=35^0\)

b: a//b

=>\(\widehat{A_4}=\widehat{B_2}\)(hai góc đồng vị)

c: \(\widehat{A_2}+\widehat{A_1}=180^0\)(hai góc kề bù)

=>\(\widehat{A_2}=180^0-35^0=145^0\)