a) Vì ABCD là tứ giác nên \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\Rightarrow\widehat{C}+\widehat{D}=360^0-90^0-60^0=210^0\)
\(\orbr{\begin{cases}\widehat{C}-\widehat{D}=20^0\\\widehat{C}+\widehat{D}=210^0\end{cases}}\Rightarrow\widehat{C}=\frac{210^0+20^0}{2}=115^0\)
\(\widehat{D}=210^0-115^0=95^0\)
Góc ngoài của C là : \(180^0-115^0=65^0\)
Tương tự câu 2 bạn làm thôi nhé
\(\widehat{C}=\frac{3}{4}\widehat{D}\Rightarrow\widehat{C}+\widehat{D}=\frac{3}{4}\widehat{D}+\widehat{D}=180^0\)
\(\Rightarrow\frac{7}{4}\widehat{D}=180^0\Leftrightarrow\widehat{D}\approx103^0\)
\(\Rightarrow\widehat{C}\approx77^0\)