ta có \(\widehat{xOz}\)và \(\widehat{y\text{O}z}\)là hai góc kề nhau
=>\(\widehat{xOz}\)+ \(\widehat{y\text{O}z}\)=\(\widehat{xOy}\)
\(b^0+\widehat{y\text{O}z}=a^0\)
\(\widehat{y\text{O}z}\)=a0-b0
Do Oz nằm giữa Ox và Oy
\(\Rightarrow\)\(\widehat{xOy}=\widehat{xOz}+\widehat{yOz}\)
\(\Leftrightarrow\)\(a=b+\widehat{yOz}\)
\(\Rightarrow\widehat{yOz}=a-b\)