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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: $116^\circ$


Team Careers360 2nd Jan, 2024
Answer (1)
Team Careers360 10th Jan, 2024

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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