Question
In $\triangle\text{ABC},$ if $\angle\text{B}=60^\circ,\angle\text{C}=80^\circ$ and the bisectors of angles $\angle\text{ABC}$ and $\angle\text{ACB}$ meet at a point O, then find the measure of $\angle\text{BOC}.$
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Answer
In $\triangle\text{ABC},$ if $\angle\text{B}=60^\circ,\angle\text{C}=80^\circ$ and the bisectors of $\angle\text{B}$ and $\angle\text{C}$ meet at O. We need to find the measure of $\angle\text{BOC}$ 
Since, BO is the bisector of $\angle\text{B}$$\angle\text{OBC}=\frac{1}{2}\angle\text{B}$
$=\frac{1}{2}(60^\circ)$
$=30^\circ$
Similarly, CO is the bisector of $\angle\text{C}$$\angle\text{OCB}=\frac{1}{2}\angle\text{C}$
$=\frac{1}{2}(80^\circ)$
$=40^\circ$
Now, applying angle sum property of the triangle, in $ \triangle\text{BOC},$ we get,$\angle\text{OCB}+ \angle\text{OBC}+ \angle\text{BOC}=180^\circ$
$30^\circ+40^\circ+ \angle\text{BOC}=180^\circ$
$\angle\text{BOC}=180^\circ-70^\circ$
$=110^\circ$
Therefore, $\angle\text{BOC}=110^\circ.$
Since, BO is the bisector of $\angle\text{B}$$\angle\text{OBC}=\frac{1}{2}\angle\text{B}$$=\frac{1}{2}(60^\circ)$
$=30^\circ$
Similarly, CO is the bisector of $\angle\text{C}$$\angle\text{OCB}=\frac{1}{2}\angle\text{C}$
$=\frac{1}{2}(80^\circ)$
$=40^\circ$
Now, applying angle sum property of the triangle, in $ \triangle\text{BOC},$ we get,$\angle\text{OCB}+ \angle\text{OBC}+ \angle\text{BOC}=180^\circ$
$30^\circ+40^\circ+ \angle\text{BOC}=180^\circ$
$\angle\text{BOC}=180^\circ-70^\circ$
$=110^\circ$
Therefore, $\angle\text{BOC}=110^\circ.$
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