Đặt \(\widehat{xOy}=a,\widehat{yOz}=b\Rightarrow a+b=180\)
a) \(\hept{\begin{cases}a+b=180\\3a+4b=640\end{cases}\Leftrightarrow}\hept{\begin{cases}a=80\\b=100\end{cases}}\)
b) \(\hept{\begin{cases}a+b=180\\a-b=20\end{cases}\Leftrightarrow\hept{\begin{cases}a=100\\b=80\end{cases}}}\)
c) \(\hept{\begin{cases}a+b=180\\2a-b=60\end{cases}\Leftrightarrow\hept{\begin{cases}a=80\\b=100\end{cases}}}\)
Bài giải
Ta có : \(\widehat{xOy}+\widehat{yOz}=180^o\)
a, Ta có :
\(3\widehat{xOy}+4\widehat{yOz}=3\left(\widehat{xOy}+\widehat{yOz}\right)+\widehat{yOz}=3\cdot180^o+\widehat{yOz}\)
\(=540^o+\widehat{yOz}=640^o\text{ }\Rightarrow\text{ }\widehat{yOz}=100^o\)
\(\Rightarrow\text{ }\widehat{xOy}=80^o\)
b, \(\widehat{xOy}= ( 180^o+20^o )\text{ : }2=100^o\)
\(\widehat{yOz}=100^o-20^o=80^o\)
c,
\(\widehat{xOy}+\widehat{yOz}=180^o\text{ }\Rightarrow\text{ }\widehat{xOy}=180^o-\widehat{yOz}\)
\(2\left(180^o-\widehat{yOz}\right)-\widehat{yOz}=360^o-3\widehat{yOz}=60^o\)
\(3\widehat{yOz}=300^o\)
\(\widehat{yOz}=100^o\)
\(\widehat{xOy}=80^o\)
