Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\widehat{A}=360^o-\widehat{B}-\widehat{C}-\widehat{D}=360^o-65^o-85^o-120^o=90^o\)
\(A=360^0-\left(B+C+D\right)=360^0-\left(65^0+85^0+120^0\right)=90^0\)
ta có :
\(A+B+C+D=360^0\Rightarrow A+B=210^0\)
mà ta có :
\(AIB=180^0-IAB-IBA=180^0-\frac{\left(A+B\right)}{2}=180^0-\frac{210^0}{2}=75^0\)
A+B+C+D=360<=> C+D=360-(A+B)=140
Ta có hpt:
\(\hept{\begin{cases}C+D=140\\C-D=20\end{cases}\Leftrightarrow\hept{\begin{cases}C=80\\D=60\end{cases}}}\)
Ta có: \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow30^0+70^0+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow\widehat{C}+\widehat{D}=260^0\left(1\right)\)
Ta lại có: \(\widehat{C}-\widehat{D}=30^0\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\widehat{C}=145^0\)
\(\Rightarrow\widehat{D}=115^0\)
Ta có : \(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=120+90+\widehat{C}+\widehat{D}=360^o\)
\(\Rightarrow\widehat{C}+\widehat{D}=150^o\)
Mà \(\widehat{C}=2\widehat{D}\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{C}=100\\\widehat{D}=50\end{matrix}\right.\)
Vậy ...
Ta có:
\(A+B+C+D=360^0\)
\(\Leftrightarrow120^0+90^0+2D+D=360^0\)
\(\Leftrightarrow3D=150^0\)
\(\Rightarrow D=50^0\)
\(C=2D=100^0\)
Xét tứ giác ABCD có:
\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^0\)
\(\Rightarrow\widehat{C}+\widehat{D}=360^0-\widehat{A}-\widehat{B}=260^0\)
Mà \(\widehat{C}-\widehat{D}=40^0\)
\(\Rightarrow\left\{{}\begin{matrix}\widehat{C}=\left(260^0+40^0\right):2=150^0\\\widehat{D}=\left(260^0-40^0\right):2=110^0\end{matrix}\right.\)
C.140
C.140