ta có\(\widehat{xOt}\)kề bù với góc \(\widehat{yOt}\)
\(\Rightarrow\widehat{xOy}=180^o\)
\(\Rightarrow\widehat{xOz}+\widehat{zOy}=180^o\)
\(\Rightarrow\widehat{xOz}=180^o-\widehat{zOy}\)
\(\Rightarrow\widehat{xOz}=180^o-80^o\)
\(\Rightarrow\widehat{xOz}=100^o\)
Vì \(\widehat{xOt}=\dfrac{1}{2}\widehat{xOz}\left(50^o=\dfrac{1}{2}.100^o\right)\)
Nên tia Ot là tia phân giác của \(\widehat{xOz}\)
Giải:
Vì \(x\widehat{O}t\) và \(y\widehat{O}t\) là 2 góc kề bù
\(\Rightarrow x\widehat{O}t+t\widehat{O}y=180^o\)
\(50^o+t\widehat{O}y=180^o\)
\(t\widehat{O}y=180^o-50^o\)
\(t\widehat{O}y=130^o\)
\(\Rightarrow y\widehat{O}z+z\widehat{O}t=y\widehat{O}t\)
\(80^o+z\widehat{O}t=130^o\)
\(z\widehat{O}t=130^o-80^o\)
\(z\widehat{O}t=50^o\)
\(\Rightarrow x\widehat{O}t+t\widehat{O}z=x\widehat{O}z\)
\(50^o+50^o=x\widehat{O}z\)
\(\Rightarrow x\widehat{O}z=100^o\)
Vì +) \(x\widehat{O}t+t\widehat{O}z=x\widehat{O}z\)
+) \(x\widehat{O}t=t\widehat{O}z=50^o\)
⇒Ot là tia p/g của \(x\widehat{O}z\)
Chúc bạn học tốt!
