Question

In the figure below, if $\Delta ABC$ is an isosceles triangle and OB and OC are angle bisectors, then $\angle BOC=$

• A. ${90}^{\circ }-\left(\frac{\angle A}{2}\right)$
• B. ${90}^{\circ }+\left(\frac{\angle A}{2}\right)$
• C. $-{90}^{\circ }-\left(\frac{\angle A}{2}\right)$
• D. $-{90}^{\circ }+\left(\frac{\angle A}{2}\right)$

Answer 1

The correct option is B

${90}^{\circ }+\left(\frac{\angle A}{2}\right)$

In $\Delta ABC$, $\angle ABC=\angle ACB$

So, $\frac{\left(\angle ABC\right)}{2}=\frac{\left(\angle ACB\right)}{2}$

$\angle BOC=\angle BCO$

Therefore, $\Delta BOC$ is also isosceles.

Also, $\angle A+\angle ABC+\angle ACB={180}^{\circ }$

Let $\angle ABC=\angle ACB=x$

So, $\angle A+x+x={180}^{\circ }$

$2x={180}^{\circ }-\angle A$

$X=\frac{{180}^{\circ }-\angle A}{2}$

$x={90}^{\circ }-\left(\frac{\angle A}{2}\right)\dots \dots \left(i\right)$

$\angle OBC=\angle OCB=\frac{x}{2}$

So, $\angle OBC+\angle OCB=x$

$\angle BOC={180}^{\circ }-(\angle OBC+\angle OCB)$

$={180}^{\circ }-x$

$={180}^{\circ }-[{90}^{\circ }-\left(\frac{\angle A}{2}\right)]$ ....... (using (i))

$={180}^{\circ }-{90}^{\circ }+\left(\frac{\angle A}{2}\right)$

$={90}^{\circ }+\left(\frac{\angle A}{2}\right)$