\(a,\widehat{tOh}=\widehat{yOh}+\widehat{tOy}=\left(\widehat{xOh}-\widehat{xOy}\right)+\dfrac{1}{2}\widehat{xOy}\\ =\left(90^0-60^0\right)+\dfrac{1}{2}\cdot60^0=30^0+30^0=60^0\)
\(b,\widehat{yOt}=\dfrac{1}{2}\widehat{xOy}=30^0=\dfrac{1}{2}\widehat{tOh}\left(30^0=\dfrac{1}{2}\cdot60^0\right)\)
Do đó Oy là phân giác \(\widehat{tOh}\)
a) \(\widehat{xOy}+\widehat{yOx'}=180^o\) (kề bù)
\(\Rightarrow\widehat{yOx'}=180^o-60^o=120^o\)
Vì \(Oh\perp Ox'\)
\(\Rightarrow\widehat{hOx'=90^o}\)
Tia Ot là phân giác của \(\widehat{yOx'}\)
\(\Rightarrow\widehat{yOt}=\widehat{tÕx'}=\dfrac{1}{2}\widehat{yOx'}=\dfrac{1}{2}.120^o=60^o\)
\(\Rightarrow\widehat{yOh}=90^o-60^o=30^o\)
\(\Rightarrow\widehat{tOh}=\widehat{yOx'}-\widehat{hOy}\)
\(\Rightarrow\widehat{tOh}=120^o-90^o=30^o\)
