a) Ta có \(\widehat{xOz}+\widehat{zOy}=\widehat{xOy}\)
\(\Leftrightarrow60^o+\widehat{zOy}=180^o\)
\(\Rightarrow\widehat{zOy}=120^o\)
b) \(\widehat{mOn}=\widehat{mOz}+\widehat{zOn}\)
\(\Leftrightarrow\widehat{mOn}=\dfrac{1}{2}\cdot\widehat{xOz}+\dfrac{1}{2}\cdot\widehat{zOy}\)
\(\Leftrightarrow\widehat{mOn}=\dfrac{1}{2}\cdot60^o+\dfrac{1}{2}\cdot120^o=30^o+60^o=90^o\)
Vậy \(\widehat{mOn}=90^o\)
Đúng 0
Bình luận (0)
