a, Vì \(\widehat{iOz}>\widehat{iOy}\left(130^o>50^o\right)\)nên tia Oy nằm giưa 2 tia còn lại
b, Vì tia Oy nằm giữa nên ta có:
\(\widehat{iOy}+\widehat{yOz}=\widehat{iOz}\)
\(50^o+\widehat{yOz}=130^o\)
\(\widehat{yOz}=130^o-50^o\)
\(\widehat{yOz}=80^o\)
c, Ta có: \(\widehat{iOz}+\widehat{iOz'}=180^o\) (kề bù)
\(130^o+\widehat{iOz'}=180^o\)
\(\widehat{iOz'}=180^o-130^o\)
\(\widehat{iOz'}=50^o\)
\(\Rightarrow\widehat{iOy}=\widehat{iOz'}\)
\(\Rightarrow\widehat{yOz'}=50^o+50^o=100^o\)
Vì \(\widehat{iOy}=\widehat{iOz'}=\dfrac{\widehat{yOz'}}{2}=\dfrac{100^o}{2}=50^o\)
nên tia Oi là tia phân giác của \(\widehat{yOz'}\)