Xét tứ giác \(ABCD\)có :
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
Mà \(\widehat{A}=36^o;\widehat{B}=48^o\)
\(\Rightarrow\widehat{C}+\widehat{D}=360^o-36^o-48^0=276^o\)
Lại có : \(\widehat{C}=2\widehat{D}\)
\(\Rightarrow2\widehat{D}+\widehat{D}=276^o\)
\(\Rightarrow3\widehat{D}=276^o\)
\(\Rightarrow\widehat{D}=92^o\)
Nên : \(\widehat{C}=92^o.2=184^o\)
Vậy \(\widehat{C}=184^o;\widehat{D}=92^o\)
Tứ giác ABCD có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)
mà \(36^o+48^o+\widehat{C}+\widehat{D}=360^o\)
=> \(84^o+\widehat{C}+\widehat{D}=360^o=>\widehat{C}+\widehat{D}=360^o-84^o=276^o\)
Tổng số phần bằng nhau là : 2 + 1 = 3 phần
\(\widehat{D}=276^o:3=92^o\)
\(\widehat{C}=276^o-92^o=184^o\)
Vậy ...
Xét tứ giác ABCD có \(\widehat{A}=36^o;\widehat{B}=48^o\)
\(\Rightarrow\widehat{C}+\widehat{D}=360^O-\widehat{A}-\widehat{B}\)
\(\Rightarrow\widehat{C}+\widehat{D}=360^O-36^O-48^O=276^O\)
Ta có: \(\widehat{C}=2\widehat{D}\Rightarrow\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}\)
Áp dụng t/c DTSBN ta có:
\(\frac{\widehat{C}}{2}=\frac{\widehat{D}}{1}=\frac{\widehat{C}+\widehat{D}}{2+1}=\frac{276^O}{3}=92^O\)
\(\Rightarrow\hept{\begin{cases}\frac{\widehat{C}}{2}=92^0\Rightarrow\widehat{C}=184^O\\\frac{\widehat{D}}{1}=92^O\Rightarrow\widehat{D}=92^O\end{cases}}\)
Vậy \(\widehat{C}=184^O;\widehat{D}=92^O\)
