Bài 3 :
Xét \(\Delta GHI\) : \(\widehat{G_2}+\widehat{H}+\widehat{I}=180^o\)
\(\Rightarrow\widehat{G_2}=180^o-\left(\widehat{H}+\widehat{I}\right)=180^o-\left(60^o+40^o\right)=80^o\)
Ta có : \(xy//HI\left(gt\right)\)
\(\Rightarrow\widehat{G_3}=\widehat{I}=40^o\)
\(\widehat{G_1}+\widehat{G_2}+\widehat{G_3}=180^o\Rightarrow\widehat{G_1}=180^o-\left(\widehat{G_2}+\widehat{G_3}\right)=180^o-\left(80^o+40^o\right)=60^o\)
Bài 2:
Ta có: Ax//By'
AB\(\perp\)Ax
Do đó: By'\(\perp\)AB
=>\(\widehat{ABC}=90^0\)
Ta có: Ax//By'
=>\(\widehat{xDC}=\widehat{DCB}\)(hai góc so le trong)
=>\(\widehat{xDC}=50^0\)
Bài 4:
Ta có: \(\widehat{tAz}=\widehat{ADC}\left(=65^0\right)\)
mà hai góc này là hai góc ở vị trí đồng vị
nên AB//DC
Ta có: AB//DC
=>\(\widehat{BCD}+\widehat{ABC}=180^0\)
mà \(\widehat{ABC}+50^0=180^0\)
nên \(\widehat{BCD}=50^0\)
