Giải
-Xét tứ giác ABCD có:
\(\left\{{}\begin{matrix}\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\\\widehat{A}=\widehat{B}=90^0\\\widehat{C}=2\widehat{D}\end{matrix}\right.\)
\(\Rightarrow90^0+90^0+2\widehat{D}+\widehat{D}=360^0\)
\(\Rightarrow3\widehat{D}=180^0\)
\(\Rightarrow\widehat{D}=60^0\)
\(\Rightarrow\widehat{C}=2\widehat{D}=2\times60^0=120^0\)
Vậy \(\widehat{D}=60^0\)
\(\widehat{C}=120^0\)
Đúng 0
Bình luận (0)
