Question : In $\triangle ABC$, the internal bisectors of $\angle ABC$ and $\angle ACB$ meet at $I$ and $\angle BAC=50°$. The measure of $\angle BIC$ is:
Option 1: $105°$
Option 2: $115°$
Option 3: $125°$
Option 4: $130°$
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Correct Answer: $115°$
Solution : Given: In $\triangle ABC$, the internal bisector of $\angle ABC$ and $\angle ACB$ meet at $I$ and $\angle BAC=50°$. $\therefore \angle ABC+\angle ACB=180°-50°$ $⇒ \angle ABC+\angle ACB=130°$ In $\triangle BIC$, $\angle IBC+\angle BIC +\angle ICB =180°$ $I$ is the internal bisector of $\angle ABC$ and $\angle ACB$. $\therefore \frac{\angle ABC}{2}+\frac{\angle ACB}{2}+\angle BIC = 180°$ $⇒\angle BIC=180°-\frac{130°}{2}$ $⇒\angle BIC=180°-65°$ $⇒\angle BIC=115°$ Hence, the correct answer is $115°$.
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