Ta có: \(\widehat{B}_{trong}+\widehat{B}_{ngoài}=180^o\) ( kề bù )
\(\Rightarrow\widehat{B}_{trong}=180^o-50^o=130^o\)
\(\Rightarrow\widehat{C}_{trong}=360^o-130^o-40^o-80^o=110^o\) ( tổng 4 góc của tứ giác )
\(\Rightarrow x=180^o-110^o=70^o\) ( kề bù )
\(\widehat{B}=180^o-50^o=130^o\\ \widehat{C}=360^o-40^o-80^o-130^o=110\\ =>x+\widehat{C}=180\left(bù\right)\\ =>x=180-110=70^o\)
ta có :
góc `B= 180^o - 50^o= 130^o`
ta có :
`130^o + 40^o + 80^o + B= 360^o`
`=> B= 360^o-( 130^o + 40^o + 80^o)`
`=> B= 360^o- 250`
`=>B= 110^o`
vậy góc `B=180^o - 110^o=70^o`
xet tu giac abcd co
\(A+B+C+D=360^o\left(DL\right)\)
hay \(40^o+\left(180-50\right)+C+80^o=360^o\)
=> \(C=110^o\)
=> \(x=180^o-110^o=70^o\)
