Ta có
\(\widehat{C1}+\widehat{C2}=180^0\) ( kề bù ) (1)
\(\widehat{C1}-\widehat{C2}=40^0\) (giả thiết ) (2)
Cộng (1) và (2)
\(\Rightarrow\left(\widehat{C1}+\widehat{C2}\right)+\left(\widehat{C1}-\widehat{C2}\right)=180^0+40^0\)
\(\Rightarrow2.\widehat{C1}=220^0\)
\(\Rightarrow\widehat{C1}=110^0\)
\(\Rightarrow\widehat{C2}=70^0\)
Mặt khác
\(\begin{cases}\widehat{C1}=\widehat{D2}\\\widehat{C1}=\widehat{D1}\end{cases}\) (a//b)
\(\Rightarrow\begin{cases}\widehat{D1}=70^0\\\widehat{D2}=110^0\end{cases}\)
Có: \(\widehat{C_1}+\widehat{C_2}=180\) (cạp góc kề bù)
=> \(\begin{cases}\widehat{C_1}+\widehat{C_2}=180\\\widehat{C_1}-\widehat{C_2}=40\end{cases}\) \(\Leftrightarrow\begin{cases}40+\widehat{C_2}+\widehat{C_2}=180\\\widehat{C_1}=40+\widehat{C_2}\end{cases}\)
\(\Leftrightarrow\begin{cases}2\widehat{C_2}=140\\\widehat{C_1}=40+\widehat{C_2}\end{cases}\)\(\Leftrightarrow\begin{cases}\widehat{C_2}=70\\\widehat{C_1}=110\end{cases}\)
=> \(\widehat{C_1}=\widehat{D_2}=110\) (cặp góc soletrong do a//b)
\(\widehat{C_2}=\widehat{D_1}=70\) (cặp góc soletrong do a//b)
Theo bài ra ta có:
\(\widehat{C1} - \widehat{C2} = 40^O\)(1)
\(\widehat{C1} + \widehat{C2} = 180^O\)(kề bù) 2
Từ (1)(2), cộng vế với vế ta có:
\(2 \widehat{C1} = 220^O\)
\(\Rightarrow\)\(\widehat{C1} = 220^O : 2 = 110^O\)
\(\Rightarrow\)\(\widehat{C2} = 110^O - 40^O = 70^O\)
Ta có:
\(\widehat{C1} = \widehat{D2}\) (so le trong)
mà \(\widehat{C1} = 110^O (cmt)\)
\(\Rightarrow\)\(\widehat{D2} = 110^O\)
\(\widehat{C2} = \widehat{D1}\)(so le trong)
mà \(\widehat{C2} = 70^O (cmt)\)
\(\Rightarrow\)\(\widehat{D1} = 70^O\)
Vậy \(\widehat{D1} = 70^O\); \(\widehat{D2} = 110^O\)
