Ta có : \(\widehat{AOB}=160^0\)
Vì \(OM\)vuông góc với \(OA\)
=> \(\widehat{AOM}=90^0\)
Vậy \(\widehat{BOM}+\widehat{MOA}=\widehat{BOA}\)
\(\Rightarrow\widehat{BOM}=\widehat{BOA}-\widehat{MOA}\)
\(\widehat{BOM}=160^0-90^0\)
\(\widehat{BOM}=70^0\)
Theo ta thấy trên góc AOM có N nằm giữa
\(\Rightarrow\widehat{AON}=\widehat{\frac{AOM}{2}}=\frac{90^0}{2}=45^0\)
Đủ điều kiện chứng minh
\(\Rightarrow\widehat{AON}+\widehat{MON}=\widehat{AOM}\)
\(\widehat{MON}=\widehat{AOM}-\widehat{AON}\)
\(\widehat{MON}=90^0-45^0\)
\(\widehat{MON}=45^0\)
Vậy ....