a) Ta có: \(\widehat{xOy}+\widehat{yOz}=180^0\)(hai góc kề bù)
\(\Leftrightarrow\widehat{zOy}+140^0=180^0\)
hay \(\widehat{yOz}=40^0\)
Vậy: \(\widehat{yOz}=40^0\)
a) Ta có: xOy+yOz=1800xOy+yOz=1800(hai góc kề bù)
⇔zOy+1400=1800
hay yOz=400yOz=400
Vậy: yOz=400