Ta có ABCD là tứ giác
=> \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
=>\(\widehat{C}+\widehat{D}=360^o-\left(\widehat{A}+\widehat{B}\right)\)
=>\(\widehat{C}+\widehat{D}=360^o-\left(120^o+50^o\right)\)
=>\(\widehat{C}+\widehat{D}=190^o\)
Lại có: \(\widehat{C}-\widehat{D}=40^o\)
=>\(\hept{\begin{cases}\widehat{C}+\widehat{D}=190^o\\\widehat{C}-\widehat{D}=40^o\end{cases}}\)
Cộng từng vế 2 phương trình, ta được :
\(2\widehat{C}=230^o\)
=>\(\widehat{C}=115^o\)
=>\(\widehat{D}=75^o\)
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