Xét tứ giác ABCD có:
$\widehat{A}$$+$$\widehat{B}$$+$$\widehat{C}$$+$$\widehat{D}$$=$$360^0$
$\Rightarrow$$\widehat{C}$$+$$\widehat{D}$$=$$360^0$$-$$\widehat{A}$$-$$\widehat{B}$$=$$260^0$
Mà $\widehat{C}$$-$$\widehat{D}$$=$$40^0$
Nên $\widehat{C}$$=(260^0+40^0):2=150^0$
$\widehat{D}$$=(260^0-40^0):2=110^0$
$\Rightarrow$$\widehat{C}$$=$$150^0$
$\Rightarrow$$\widehat{D}$$=$$110^0$
@Nobitao@