=
â ) Ta có : \(\widehat{AOC}=\widehat{AOB}+\widehat{BOC}=160^o\)( 2 góc kề có tổng = 160 \(^o\) gt )
Ta có : \(\widehat{AOB}=\left(\widehat{AOC}+\widehat{AOB}-\widehat{BOC}\right):2\)
\(\widehat{AOB}=\left(160^o+120^o\right):2\)
\(\widehat{AOB}=140^O\)
Ta có : \(\widehat{BOC}=\widehat{AOC}-\widehat{AOB}\)
\(\widehat{BOC}=160^o-140^o=120^o\)
b)
c)ta co : \(\widehat{C'OC}\)là góc bẹt
=> \(\widehat{C'OC}\) = \(180^o\)
Ta có : \(\widehat{BOC'}=\widehat{C'OC}-\widehat{BOC}\) ( 2 GÓC KỀ NHAU )
\(\widehat{BOC'}=180^O-20^O=160^O\) ( 1 )
ma : \(\widehat{AOC}=160^O\)( chứng minh trên câu a ) ( 2 )
Từ ( 1 ) vả ( 2 ) suy ra \(\widehat{BOC'}=\widehat{AOC}=160^O\)
CHÚC BẠN HỌC TỐT !!!
