AB//CD
nên \(\widehat{B}+\widehat{C}=180^0\)
mà \(\widehat{B}-\widehat{C}=70^0\)
nên \(\widehat{B}=\dfrac{180^0+70^0}{2}=125^0\)
\(\widehat{C}=180^0-\widehat{B}=180^0-125^0=55^0\)
Ta có AB//CD
nên \(\widehat{D}+\widehat{A}=180^0\)
\(\Leftrightarrow\widehat{A}\cdot\dfrac{2}{7}+\widehat{A}=180^0\)
\(\Leftrightarrow\widehat{A}=180^0:\dfrac{9}{7}=140^0\)
\(\widehat{D}=\dfrac{2}{7}\cdot140^0=40^0\)