VD4:
\(Ta.có:\widehat{xOy}+\widehat{yOz}=180^o\\ \Leftrightarrow50^{^o}+\widehat{yOz}=180^o\\ \Rightarrow\widehat{yOz}=180^o-50^o=130^o\\ Ta.có:\widehat{yOt}=\widehat{tOz}=\dfrac{1}{2}\widehat{yOz}\left(t.c.tia.phân.giác\right)\\ \Leftrightarrow\widehat{yOt}=\widehat{tOz}=\dfrac{130^o}{2}=65^o\\ Vậy:\widehat{xOt}=\widehat{xOy}+\widehat{yOt}=50^o+65^o=115^o\\ Chọn.B\)
VD5:
\(Ta.có:\widehat{xOy}+\widehat{yOz}=180^o\left(2.góc.kề.bù\right)\\ Lại.có:\widehat{mOn}=\widehat{mOy}+\widehat{yOn}\\ Mà:\widehat{xOm}=\widehat{mOy}=\dfrac{1}{2}\widehat{xOy};\widehat{yOn}=\widehat{nOz}=\dfrac{1}{2}\widehat{yOz}\left(t.c.tia.phân.giác\right)\\ \Rightarrow\widehat{mOn}=\dfrac{1}{2}\left(\widehat{xOy}+\widehat{yOz}\right)=\dfrac{1}{2}.180^o=90^o\\ Chọn.C\)
