Bài 4:
$AB∩CD≡O$
⇒ $\widehat{AOD}=\widehat{BOC}=110^o$ (đối đỉnh)
Ta có: $\widehat{AOD}+\widehat{BOD}=180^o$
$110^o+\widehat{BOD}=180^o$
⇒ $\widehat{BOD}=180^o-110^o=70^o$
⇒ $\widehat{BOD}=\widehat{AOC}=70^o$
Vậy $\widehat{BOC}=110^o$, $\widehat{BOD}=\widehat{AOC}=70^o$
Bài 5:
Ta có: $\widehat{AOC}-\widehat{AOD}=20^o$
⇒ $\widehat{AOC}=20^o+\widehat{AOD}$
mà $\widehat{AOC}+\widehat{AOD}=180^o$
⇒ $20^o+\widehat{AOD}+\widehat{AOD}=180^o$
$20^o+2.\widehat{AOD}=180^o$
$2.\widehat{AOD}=180^o-20^o=160^o$
⇒ $\widehat{AOD}=160^o:2=80^o$
⇒ $\widehat{AOC}=80^o+20^o=100^o$
$AB∩CD≡O$
⇒ $\widehat{AOC}=\widehat{BOD}=100^o$ (đối đỉnh)
$\widehat{AOD}=\widehat{BOC}=80^o$ (đối đỉnh)
Vậy $\widehat{AOC}=\widehat{BOD}=100^o$ (đối đỉnh)
$\widehat{AOD}=\widehat{BOC}=80^o$ (đối đỉnh)
