Ta có hình vẽ:
a/ Ta có: \(\widehat{xOz}+\widehat{zOy}=180^0\left(kb\right)\)
hay \(\widehat{xOz}+\dfrac{2}{3}.\widehat{xOz}=180^0\)
hay \(\dfrac{5}{3}.\widehat{xOz}=180^0\)
\(\Rightarrow\widehat{xOz}=180^0:\dfrac{5}{3}=108^0\)
Ta có: \(\widehat{zOy}=\dfrac{2}{3}.\widehat{xOz}\)
hay \(\widehat{zOy}=\dfrac{2}{3}.108^0=72^0\)
b/ Ta có: \(\widehat{mOz}=\dfrac{1}{2}.\widehat{xOz}\)
và \(\widehat{zOn}=\dfrac{1}{2}.\widehat{zOy}\)
==> \(\widehat{mOz}+\widehat{zOn}=\dfrac{1}{2}.\left(\widehat{xOz}+\widehat{zOy}\right)\)
hay \(\widehat{mOn}=\dfrac{1}{2}.\widehat{xOy}\)
\(\widehat{mOn}=\dfrac{1}{2}.180^0=90^0\)
Vậy góc zOm và góc zOn phụ nhau.
