Vẽ Oz // Ax
=> \(\widehat{xAO}=\widehat{AOz}\left(soletrong\right)=50^0\)
Ta có: \(\widehat{AOz}+\widehat{zOB}=\widehat{AOB}\)
hay \(50^0+\widehat{zOB}=80^0\)
\(\widehat{zOB}=80^0-50^0\)
=> \(\widehat{zOB}=30^0\)
Vì Ax // Oz
mà Ax // By
=> Oz // By
=> \(\widehat{OBy}=\widehat{zOB}\left(soletrong\right)=30^0\)
=> \(\widehat{OBy}=30^0\)