Xét tứ giác ABCD có
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Leftrightarrow\widehat{C}+\widehat{D}=240^0\)
mà \(\widehat{C}-\widehat{D}=62^0\)
nên \(2\cdot\widehat{C}=302^0\)
\(\Leftrightarrow\widehat{C}=151^0\)
\(\Leftrightarrow\widehat{D}=89^0\)