In $$\triangle ABC, BO$$ and $$CO$$ are the bisectors of the $$\angle ABC$$ and $$\angle ACB$$. If the measure of $$\angle A = 54^\circ$$, what is the measure of $$\angle BOC$$
$$\angle A$$+$$\angle B$$+$$\angle C$$ = 180
Or,Ā $$\angle B$$+$$\angle C$$ = 126
Now, the angle bisectors divideĀ $$\angle B$$ andĀ $$\angle C$$ into half, hence their sum is 63
Now, inĀ $$\triangle BOC$$, we 63+$$\angle BOC$$ = 180, orĀ $$\angle BOC$$ = $$117^\circ$$