Question

$O$ is the incentre of $△ABC$ and $\angle BOC={116}^{\circ }$. Then $\angle BAC=$

• A. ${75}^{\circ }$
• B. ${55}^{\circ }$
• C. ${52}^{\circ }$
• D. ${60}^{\circ }$

Answer 1

The correct option is C

${52}^{\circ }$

In $△ABC,\angle A+\angle B+\angle C={180}^{\circ }$ [Angle sum property]

$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C={90}^{\circ }$

$\Rightarrow \frac{1}{2}\angle BAC+\angle OBC+\angle OCB={90}^{\circ }$ [Since, $O$ is incentre, so, $OB$ is the angle bisector of$\angle ABC$ and $OC$ is the angle bisector of $\angle ACB$

$\Rightarrow \angle OBC+\angle OCB$ = ${90}^{\circ }$ - $\frac{1}{2}\angle BAC$ ... (i)

In $△OBC,\angle OBC+\angle OCB+\angle BOC={180}^{\circ }$ [Angle sum property]

$\Rightarrow \angle BOC$ + ${90}^{\circ }$ - $\frac{1}{2}\angle BAC$ = ${180}^{\circ }$ [Using (i)]

$\Rightarrow \angle BOC$ = ${90}^{\circ }$ + $\frac{1}{2}\angle BAC$ ... (ii)

Given, $\angle BOC$ = ${116}^{\circ }$

$\Rightarrow \angle BAC$ = $2$( $\angle BOC$ - ${90}^{\circ }$) [From (ii)]

$\Rightarrow \angle BAC$ = $2$ ( ${116}^{\circ }$ - ${90}^{\circ }$ )

$\Rightarrow \angle BAC$ = ${52}^{\circ }$