Question
In $\triangle\text{ABC},$ if bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ intersect at $O$ at angle of $120^\circ ,$ then find the measure of $\angle\text{A}.$
✓
Answer
In the given $\triangle\text{ABC},\angle\text{ABC}=\angle\text{ACB},$ the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at $O$ and $\angle\text{BOC}=120^\circ$ We need to find the measure of $\angle\text{A}$ 
So here, using the corollary, "if the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ of a $\triangle\text{ABC},$
meet at a point $O,$ Then $\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}"$
Thus, in $\triangle\text{ABC},$
$\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ-90^\circ=\frac{1}{2}\angle\text{A}$
$\angle\text{A}=2(30^\circ)$
$\angle\text{A}=60^\circ$ Thus, $\angle\text{A}=60^\circ$
So here, using the corollary, "if the bisectors of $\angle\text{ABC}$ and $\angle\text{ACB}$ of a $\triangle\text{ABC},$
meet at a point $O,$ Then $\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}"$
Thus, in $\triangle\text{ABC},$
$\angle\text{BOC}=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ=90^\circ+\frac{1}{2}\angle\text{A}$
$120^\circ-90^\circ=\frac{1}{2}\angle\text{A}$
$\angle\text{A}=2(30^\circ)$
$\angle\text{A}=60^\circ$ Thus, $\angle\text{A}=60^\circ$
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