Question : In a $\triangle \mathrm{ABC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ meet at $\mathrm{O}$. If $\angle \mathrm{BOC}=142^{\circ}$, then the measure of $\angle \mathrm{A}$ is:
Option 1: $52^\circ$
Option 2: $68^\circ$
Option 3: $104^\circ$
Option 4: $116^\circ$
Correct Answer: $104^\circ$
Solution : Given: $\angle BOC=142^\circ$ In $\triangle BOC$, ⇒ $\angle BOC+\angle OBC+\angle OCB=180^\circ$ ($\because$ OB and OC bisect $\angle B$ and $\angle C$ respectively) ⇒ $\angle BOC+\frac{1}{2}\angle B+\frac{1}{2}\angle C=180^\circ$ ⇒ $\angle BOC=180^\circ-\frac{1}{2}(\angle B+\angle C)$ ⇒ $\angle BOC=180^\circ-\frac{1}{2}(180^\circ-\angle A)$ ⇒ $\angle BOC=90^\circ+\frac{1}{2}\angle A$ ⇒ $142^\circ=90^\circ+\frac{1}{2}\angle A$ ⇒ $52^\circ=\frac{1}{2}\angle A$ ⇒ $\angle A=104^\circ$ Hence, the correct answer is $104^\circ$.
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