Ta có: Trong tứ giác \(ABCD\): \(\widehat A+\widehat B+\widehat C+\widehat D=360^0\)
Thay \(\widehat B =\widehat A+10^0,\)\(\widehat C=\widehat B+10^0,\)\(\widehat D=\widehat C+10^0\) vào ta được:\(\widehat A+\widehat A+10^0+\widehat B+10^0+\widehat C+10^0=360^0\)
\(\Rightarrow 2\widehat A+\widehat B+\widehat C=330^0\)
\(\Rightarrow 2\widehat A+\widehat A+10^0+\widehat B+10^0=330^0\)
\(\Rightarrow 3\widehat A+\widehat B=310^0\)
\(\Rightarrow 3\widehat A+\widehat A+10^0=310^0\)
\(\Rightarrow 4\widehat A=300^0\)
\(\Rightarrow \widehat A=75^0\)
\(\Rightarrow \widehat B=85^0\)
Vậy chọn \(B.\) \(\widehat B = {85^0}\)
