a) Do OC là tia phân giác góc AOB nên $\widehat{BOC}=\widehat{AOC}=\frac{1}{2}\widehat{AOB}=\frac{1}{2}60^{\circ}=30^{\circ}$

Ta có: $\widehat{BOE}+\widehat{AOB}=180^{\circ}$;$\widehat{COE}+\widehat{AOC}=180^{\circ};\widehat{AOC}+\widehat{AOD}=180^{\circ}$4 (các cặp góc kề bù) nên

$\widehat{BOE}=180−\widehat{AOD}=180^{\circ}−60^{\circ}=120^{\circ}$

$\widehat{COE}=\widehat{AOD}=180^{\circ}−\widehat{AOC}=180^{\circ}−30^{\circ}=150^{\circ}$

b) Ta có $\widehat{BOD}+\widehat{BOC}=180^{\circ}$ (hai góc kề bù )nên

$\widehat{BOD}=180^{\circ}−\widehat{BOC}=180^{\circ}−30^{\circ}=150^{\circ}$ (hai góc kề bù ) 

Do đó $\widehat{AOD}=\widehat{BOD}$ (cùng bằng $150^{\circ}$)