Question

The bisectors of $\angle B$ and $\angle C$ of a triangle $ABC$ meet at a point $O.$ If $\angle A={60}^{\circ },$ then $\angle BOC$ is :

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is D ${120}^{\circ }$

REF.Image

$OB$ & $OC$ are angle bisector of $\angle B$ & $\angle C$

In $\Delta BOC$

$\angle 1+\angle 2+\angle BOC={180}^{\circ }$ _________ (1)

(sum of all Angles of $\Delta$)

$eq\left(1\right)\times 2$

$2\angle 2+2\angle 2+2\angle BOC={360}^{\circ }$ $\because \angle ABC=2\angle 1$

$\angle ABC+\angle ACB+2\angle BOC={360}^{\circ }$ $\angle ACB=2\angle 2$

In $\Delta ABC$

$\angle ABC+\angle ACB+\angle BAC={180}^{\circ }$ ____________ (2)

eq. (2)-(3)

$2\angle BOC-\angle BAC={180}^{\circ }$ given $\angle BAC={60}^{\circ }$

$2\angle BOC-\angle BAC=180+60$

$2\angle BOC=240$

$\angle BOC={120}^{\circ }$