Giúp em bài này vs ạ, em cảm ơn ạ

$a)$ Vì $ABCD$ nội tiếp $(O)$

`->` $\widehat{ADC}+\widehat{ABC}=180^o$

`->` $40^o+\widehat{ABC}=180^o$

`->` $\widehat{ABC}=140^o$

Tương tự ta có: $\widehat{BAD}+\widehat{BCD}=180^o$

`->` $\widehat{BAD}+100^o=180^o$

`->` $\widehat{BAD}=80^o$

$b)$ Ta có: $\widehat{CBX}+\widehat{ABC}=180^o$ (kề bù)

`->` $\widehat{CBX}+140^o=180^o$

`->` $\widehat{CBX}=80^o$

Tương tự ta có: $\widehat{BCX}+\widehat{BCD}=180^o$ (kề bù)

`->` $\widehat{BCX}+100^o=180^o$

 `->` $\widehat{BCX}=80^o$

`->` $\widehat{BXC}=180^o-\widehat{BCX}-\widehat{CBX}=180^o-80^o-40^o=60^o$