Ta có: \(\widehat{yAB}+\widehat{BAC}=180^0\left(kề-bù\right)\)
\(\Rightarrow\widehat{BAC}=180^0-\widehat{yAB}=180^0-100^0=80^0\)
Xét \(\Delta ABC\) có:
\(\widehat{B}+\widehat{A1}+\widehat{C}=180^0\left(Tổng-3-góc-trong-1\Delta\right)\)
\(\Rightarrow\widehat{C}=180^0-\widehat{A1}-\widehat{B}=180^0-80^0-70^0\)
\(\Rightarrow\widehat{C}=30^0\)
\(\Rightarrow x=30^0\)