O a b c t
a) Vì \(\widehat{aOb}< \widehat{cOa}\) (\(30^o< 110^o\))
=> Ob nằm giữa Oa, Oc
=> \(\widehat{aOb}+\widehat{bOc}=\widehat{aOc}\)
mà \(\widehat{aOb}=30^o,\widehat{aOc}=110^o\)
=> \(\widehat{bOc}=80^o\)
b) Vì Ot là tia phân giác của \(\widehat{bOc}\)
=> \(\widehat{tOb}=\dfrac{\widehat{bOc}}{2}=\dfrac{80^o}{2}=40^o\)
Ta có: Ob nằm giữa Ot, Oa
=> \(\widehat{aOb}+\widehat{bOt}=\widehat{aOt}\)
mà \(\widehat{aOb}=30^o,\widehat{tOb}=40^o\)
=> \(\widehat{aOt}=70^o\)
c) Vì \(\widehat{aOb}e\widehat{bOt}\left(30^oe40^o\right)\)
=> Ob không phải tia phân giác \(\widehat{aOt}\)
