\(\widehat{O_2}+\widehat{O_3}=180^0\left(kề.bù\right)\\ \widehat{O_3}-\widehat{O_2}=18^0\\ \Rightarrow\widehat{O_3}=\left(180^0+18^0\right):2=99^0\\ \Rightarrow\widehat{O_2}=180^0-99^0=81^0\\ \left\{{}\begin{matrix}\widehat{O_2}=\widehat{O_4}=81^0\left(đối.đỉnh\right)\\\widehat{O_3}=\widehat{O_1}=99^0\left(đối.đỉnh\right)\end{matrix}\right.\)
Vì \(\widehat{LOP}và\widehat{POM}\) là 2 góc kề bù
Và \(\widehat{POL}và\widehat{MON};\widehat{POM}và\widehat{LON}\) là 2 góc đối đỉnh
\(\Rightarrow\widehat{O4}\) = \(\left(180^o-18^o\right):2=81^{^o}\)
\(\Rightarrow180^o-81^o=99^o\)
