Xét tứ giác EFGH ta có:
\(\widehat{E}+\widehat{F}+\widehat{G}+\widehat{H}=360^o\) (theo định lý tổng các góc trong tứ giác)
\(\Rightarrow\widehat{G}+\widehat{H}=360^o-\widehat{E}-\widehat{F}\)
\(\Rightarrow\widehat{G}+\widehat{H}=360^o-70^o-80^o=210^o\)
Theo bài ra:
\(\widehat{G}-\widehat{H}=20^o\)
\(\Rightarrow\left(\widehat{G}+\widehat{H}\right)+\left(\widehat{G}-\widehat{H}\right)=210^o-20^o\)
\(\Rightarrow2\widehat{G}=190^o\Rightarrow\widehat{G}=95^o\)
\(\Rightarrow\widehat{H}=210^o-\widehat{G}=210^o-95^o=115^o\)
Vậy.....
Chúc bạn học tốt!!!
Ta có :
\(\widehat{E}+\widehat{F}+\widehat{G}+\widehat{H}=360^0\)
\(\Leftrightarrow70^0+80^0+\widehat{G}+\widehat{H}=360^0=>\widehat{G}+\widehat{H}=210^0=>\widehat{G}=210^0-H\left(1\right)\)
Mà theo đề \(\widehat{G}-\widehat{H}=20^0\)(2)
Thay (1) vào (2) \(210^0-2\widehat{H}=20^0=>\widehat{H}=\dfrac{210-20}{2}=95^0=>\widehat{G}=115^0\)
