Question : Internal bisectors of $\angle$ B and $\angle$ C of $\triangle$ ABC meet at O. If $\angle$ BAC = $80^{\circ}$, then the value of $\angle$ BOC is:
Option 1: $120^{\circ}$
Option 2: $140^{\circ}$
Option 3: $110^{\circ}$
Option 4: $130^{\circ}$
Correct Answer: $130^{\circ}$
Solution :
OB and OC are the angle bisectors of $\angle B$ and $\angle C$.
In $\triangle OBC$,
$\angle OBC + \angle BCO + \angle COB = 180^{\circ}$
$⇒\frac{1}{2} \angle B + \frac{1}{2} \angle C + \angle COB = 180^{\circ}$
$⇒\frac{1}{2} (\angle B+\angle C) + \angle COB = 180^{\circ}$
$⇒\frac{1}{2} (180^{\circ}-\angle A) + \angle COB = 180^{\circ}$
$⇒\angle COB = 180^{\circ} – (\frac{1}{2} (180^{\circ}-80^{\circ}))= 180^{\circ}$ – $50^{\circ}= 130^{\circ}$
Hence, the correct answer is $130^{\circ}$.
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