a) ta có \(\widehat{AOC}=\widehat{BOC}=\dfrac{1}{2}\widehat{AOB}=\dfrac{144}{2}=72^o\) (\(OC\) là tia phân giác \(\widehat{AOB}\))
ta có : \(\widehat{MOC=\widehat{CON}}=72-20=52^o\) \(\left(\widehat{AOM}=\widehat{BON}=20^o\right)\)
\(\Rightarrow\) \(OC\) là tia phân giác của \(\widehat{MON}\) \(\left(\widehat{MOC}=\widehat{CON}=52^o\right)\)(ĐPCM)
b) ta có \(\widehat{AOB'}=\widehat{B'OB}-\widehat{AOB}=180-144=36^o\)
ta có : \(\widehat{AOC}=\widehat{BOC}=72^o\) (chứng minh trên)
\(\Rightarrow\) \(\widehat{AOB'}< \widehat{AOC}=\widehat{BOC}\)