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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