Question : In $\triangle A B C$ the internal bisectors of $\angle ABC$ and $\angle ACB$ meet at $X$ and $\angle BAC=30^{\circ}$. The measure of $\angle BXC $ is:
Option 1: $120^{\circ}$
Option 2: $115^{\circ}$
Option 3: $105^{\circ}$
Option 4: $150^{\circ}$
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Correct Answer: $105^{\circ}$
Solution :
Given: $\angle \mathrm{BAC}=30^{\circ}$
We know that if the internal bisectors of $\angle ABC$ and $\angle ACB$ meet at $\mathrm{X}$ in $\triangle ABC$. Then,
$\angle BXC=90^{\circ}+\frac{1}{2}\angle BAC$
⇒ $\angle BXC=90^{\circ}+\frac{1}{2}×30^{\circ}$
⇒ $\angle BXC=105^{\circ}$
Hence, the correct answer is $105^{\circ}$.
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