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I is the incenter of a triangle $$ABC$$ in which $$\angle A = 80^\circ$$. $$\angle BIC =$$
The incenter satisfies the standard identity $$\angle BIC = 90^\circ + \frac{\angle A}{2}$$, so $$\angle BIC = 90^\circ + 40^\circ = 130^\circ$$.
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