@nan
$\text{bài 0.4 }$
$\text{AB và CD cắt nhau tại O }$
$\text{ => $\widehat{AOD}$ = $\widehat{COD}$ = $110^{0}$ ( 2 góc đối đỉnh )} $
$\text{có : $\widehat{AOC}$ + $\widehat{AOD}$ = $180^{0}$ ( 2 góc kề bù ) }$
$\text{=> $\widehat{AOC}$ = $180^{0}$ - $\widehat{AOD}$ = $180^{0}$ - $110^{0}$ = $70^{0}$ }$
$\text{suy ra $\widehat{DOB}$ = $\widehat{AOC}$ = $70^{0}$ ( 2 góc đối đỉnh ) }$
_________________________________________
$\text{bài 0.8 }$
$\text{Do OC nằm giữa OA và OB }$
$\text{nên suy ra $\widehat{AOC}$ + $\widehat{BOC}$ = $\widehat{AOB}$}$
$\text{⇔ $\widehat{BOC}$ = $\widehat{AOB}$ - $\widehat{AOC}$ }$
$\text{= $120^{0}$ - $30^{0}$ }$
$\text{= $90^{0}$ }$
$\text{suy ra OB ⊥ OC }$
