\(a,\left\{{}\begin{matrix}m//n\\n\perp a\end{matrix}\right.\Rightarrow m\perp a\)(từ vuông góc đến song song)
\(b,m//n\Rightarrow\widehat{N_1}=\widehat{M_1}=110^0\left(đồng.vị\right)\)
Ta có \(\widehat{M_1}+\widehat{M_2}=180^0\left(kề.bù\right)\Rightarrow\widehat{M_2}=180^0-110^0=70^0\)
Ta có \(\widehat{M_2}=\widehat{M_3}=70^0\left(đối.đỉnh\right)\)
