Question

The internal bisectors of the angles B and C of a triangle ABC meet at O. Then, $\angle BOC$ is equal to

• A. ${90}^0+A$
• B. $2A$
• C. ${90}^0+\frac{1}{2}A$
• D. ${180}^0-A$

Answer 1

The correct option is C ${90}^0+\frac{1}{2}A$

Given: OB and OC bisect $\angle B$ and $\angle C$ respectively.
In $△BOC$,
$\angle BOC+\angle OBC+\angle OCB=180$ (OB and OC bisect $\angle B$ and $\angle C$ respectively)
$\angle BOC+\frac{1}{2}\angle B+\frac{1}{2}\angle C=180$
$\angle BOC=180-\frac{1}{2}(\angle B+\angle C)$
$\angle BOC=180-\frac{1}{2}(180-\angle A)$ (Angle sum property)
$\angle BOC=90+\frac{1}{2}\angle A$