a,Ta có: \(\widehat{AOB}+\widehat{BOC}=160^o\left(1\right)\)
\(\widehat{AOB}-\widehat{BOC}=120^o\left(2\right)\)
Lấy (1)+(2): ta có:
\(2\widehat{AOB}=280^o\Rightarrow\widehat{AOB}=140^o\Rightarrow\widehat{BOC}=20^o\)
b, Ta có: \(\widehat{COB}+\widehat{BOD}=90^o\)
\(\Rightarrow\widehat{BOD}=90^o-\widehat{COB}=90^0-20^o=70^o\left(1\right)\)
Chứng minh tương tự ta có: \(\widehat{DOA}=70^o\left(2\right)\)
Từ (1);(2); Suy ra:
OD là tia phân giác góc BOA
c, Ta có: \(\widehat{AOC}=160^o\left(gt\right)\)
Lại có: \(\widehat{COD}+\widehat{C'OD}=180^o\)
\(\widehat{C'OD}=180^o-90^o=90^o\)
\(\widehat{C'OA}+\widehat{AOD}=90^o\Rightarrow\widehat{C'OA}=90^o-\widehat{AOD}=90^o-70^o=20^o\)\(\)
\(\widehat{BOC'}=\widehat{BOA}+\widehat{AOC'}=140^o+20^o=160^o\)
