Giải
a) Vì \(\left\{{}\begin{matrix}Oz\perp Ox\left(gt\right)\\Ot\perp Oy\left(gt\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{xOz}=90^o\\\widehat{yOt}=90^o\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\widehat{xOt}+\widehat{zOt}=90^o\left(2gockenhau\right)\\\widehat{yOz}+\widehat{zOt}=90^o\left(2gockenhau\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{xOt}+\widehat{zOt}+\widehat{yOz}+\widehat{zOt}=90^o+90^o=180^o\left(1\right)\)
Ta có: \(\widehat{xOt}+\widehat{zOt}+\widehat{yOz}=\widehat{xOy}=120^o\left(2\right)\)
Từ \(\left(1\right)\&\left(2\right)\Rightarrow120^o+\widehat{zOt}=180^o\)
\(\Rightarrow\widehat{zOt}=60^o\)
b) Ta có: \(\left\{{}\begin{matrix}\widehat{xOt}+\widehat{zOt}=90^o\left(cmt\right)\\\widehat{yOz}+\widehat{zOt}=90^o\left(cmt\right)\end{matrix}\right.\)
\(\Rightarrow\widehat{xOt}=\widehat{yOz}=90^o-\widehat{zOt}=90^o-60^o=30^o\)
\(\Rightarrow\frac{1}{2}.\widehat{xOt}=\frac{1}{2}.\widehat{yOz}=\frac{1}{2}.30^o\)
\(\Leftrightarrow\widehat{mOt}=\widehat{nOz}=15^o\)
Ta có: \(\widehat{mOn}=\widehat{mOt}+\widehat{zOt}+\widehat{nOz}\)
\(\Leftrightarrow\widehat{mOn}=15^o+60^o+15^o\)
\(\Leftrightarrow\widehat{mOn}=90^o\)
\(\Rightarrow Om\perp On\)
