[tex]\it m(\widehat{COB}) =m(\widehat{AOB}) -m(\widehat{AOC}) =84^o-18^o=66^o\\ \\ \widehat{COD} \equiv\widehat{DOB} \Rightarrow [OD\ -\ bisectoare\ pentru\ \widehat{COB} \Rightarrow m(\widehat{DOB}) =\dfrac{m(\widehat{COB})}{2}=\\ \\ \\ =\dfrac{66^o}{2}=33^o[/tex]