O 100* y x x' a)
Ta có \(\widehat{xOy}+\widehat{yOx'}=180^0\) (kề bù)
Thay số: 1000+ \(\widehat{yOx'}=180^0\)
\(\Rightarrow\widehat{yOx'}=180^0-100^0\)
\(\Rightarrow\widehat{yOx'}=80^0\)
Vậy \(\widehat{yOx'}=80^0\)
b)
Ot là tia phân giác của \(\widehat{xOy}\)
\(\Rightarrow\widehat{xOt}=\widehat{yOt}=\dfrac{\widehat{xOy}}{2}\)
Thay số: \(\widehat{xOt}=\widehat{yOt}=\dfrac{100^0}{2}=50^0\)
Ta lại thấy \(\widehat{xOt}+\widehat{tOx'}=180^0\) (kề bù)
Thay số: 500+ \(\widehat{tOx'}\) = 1800
=> \(\widehat{tOx'}\) = 1800- 500= 1300
Vậy \(\widehat{tOx'}=130^0\)
