\(a,\left\{{}\begin{matrix}\widehat{AON}+\widehat{AOM}=\widehat{MON}\\\widehat{BOM}+\widehat{BON}=\widehat{MON}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\widehat{AON}=120^0-90^0=30^0\\\widehat{BOM}=120^0-90^0=30^0\end{matrix}\right.\\ \Rightarrow\widehat{AON}=\widehat{BOM}\)
\(b,\widehat{xOy}=\widehat{xOA}+\widehat{AOB}+\widehat{BOy}\\ =\dfrac{1}{2}\widehat{AON}+\widehat{AOB}+\dfrac{1}{2}\widehat{BOM}\\ =15^0+15^0+\left(\widehat{AOM}-\widehat{BOM}\right)=30^0+\left(90^0-30^0\right)=90^0\\ \Rightarrow Ox\perp Oy\)
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