Question

 Cho AB//DE. Tính CAB

Answer 1

Kẻ Cx//DE//AB(Cx và AB nằm ở hai nửa mặt phẳng đối nhau bờ AC)

Cx//DE

=>\(\widehat{xCD}=\widehat{CDE}\)

=>\(\widehat{xCD}=50^0\)

Ta có: \(\widehat{xCD}+\widehat{xCA}=\widehat{ACD}\)

=>\(\widehat{xCA}+50^0=120^0\)

=>\(\widehat{xCA}=70^0\)

Ta có: Cx//AB

=>\(\widehat{CAB}=\widehat{xCA}\)(hai góc so le trong)

mà \(\widehat{xCA}=70^0\)

nên \(\widehat{CAB}=70^0\)