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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, $$\angle B$$+$$\angle C$$ = 126
Now, the angle bisectors divide $$\angle B$$ and $$\angle C$$ into half, hence their sum is 63
Now, in $$\triangle BOC$$, we 63+$$\angle BOC$$ = 180, or $$\angle BOC$$ = $$117^\circ$$
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