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: $\widehat{BOE}+\widehat{AOB}=180^{\circ}; \widehat{COE}+\widehat{AOC}=180^{\circ}; \widehat{AOC}+\widehat{AOD}=180^{\circ}$ (các cặp góc kề bù) nên
$\widehat{BOE}=180-\widehat{AOB}=180^{\circ}-60^{\circ}=120^{\circ}$
$\widehat{COE}=\widehat{AOD}=180^{\circ}-\widehat{AOC}=180^{\circ}-30^{\circ}=150^{\circ}$
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 $150^{\circ}$)