分类讨论：①当点$B$和点$C$在$AO$两侧时，过点$O$作$OP\bot AB$于点$P$，作$OQ\bot AC$于点$Q$，如图，

$\therefore AP=\frac{1}{2}AB=\sqrt{2}$.
$\because OA=2$，
$\therefore OP=\sqrt{O{A^2}-A{P^2}}=\sqrt{{2^2}-{{(\sqrt{2})}^2}}=\sqrt{2}$，
$\therefore AP=OP$，
$\therefore \angle PAO=45^{\circ}$.
$\because AQ=\frac{1}{2}AC=\sqrt{3}$，$OA=2$，
$\therefore OQ=\sqrt{O{A^2}-A{Q^2}}=\sqrt{{2^2}-{{(\sqrt{3})}^2}}=1$，
$\therefore OQ=\frac{1}{2}OA$，
$\therefore \angle QAO=30^{\circ}$，
$\therefore \angle BAC=\angle PAO+\angle QAO=75^{\circ}$
$\therefore \angle BOC=2\angle BAC=150^{\circ}$；
②当点$B$和点$C$在$AO$同侧时，过点$O$作$OM\bot AB$于点$M$，作$ON\bot AC$于点$N$，如图，

由①同理可得：$\angle MAO=45^{\circ}$，$\angle NAO=30^{\circ}$，
$\therefore \angle BAC=\angle MAO-\angle NAO=15^{\circ}$，
$\therefore \angle BOC=2\angle BAC=30^{\circ}$.
综上可知$\angle BOC$的度数为$150^{\circ}$或$30^{\circ}$.
故答案为：$150^{\circ}$或$30^{\circ}$.

分类讨论：①当点$B$和点$C$在$AO$两侧时，过点$O$作$OP\bot AB$于点$P$，作$OQ\bot AC$于点$Q$，如图，

$\therefore AP=\frac{1}{2}AB=\sqrt{2}$.
$\because OA=2$，
$\therefore OP=\sqrt{O{A^2}-A{P^2}}=\sqrt{{2^2}-{{(\sqrt{2})}^2}}=\sqrt{2}$，
$\therefore AP=OP$，
$\therefore \angle PAO=45^{\circ}$.
$\because AQ=\frac{1}{2}AC=\sqrt{3}$，$OA=2$，
$\therefore OQ=\sqrt{O{A^2}-A{Q^2}}=\sqrt{{2^2}-{{(\sqrt{3})}^2}}=1$，
$\therefore OQ=\frac{1}{2}OA$，
$\therefore \angle QAO=30^{\circ}$，
$\therefore \angle BAC=\angle PAO+\angle QAO=75^{\circ}$
$\therefore \angle BOC=2\angle BAC=150^{\circ}$；
②当点$B$和点$C$在$AO$同侧时，过点$O$作$OM\bot AB$于点$M$，作$ON\bot AC$于点$N$，如图，

由①同理可得：$\angle MAO=45^{\circ}$，$\angle NAO=30^{\circ}$，
$\therefore \angle BAC=\angle MAO-\angle NAO=15^{\circ}$，
$\therefore \angle BOC=2\angle BAC=30^{\circ}$.
综上可知$\angle BOC$的度数为$150^{\circ}$或$30^{\circ}$.
故答案为：$150^{\circ}$或$30^{\circ}$.