Question

In a $\Delta ABC,$ if $\angle A={60}^{\circ },\angle B={80}^{\circ }$ and the bisectors of $\angle B$ and $\angle C$ meet at $O$, then $\angle BOC=$___

Answer 1

Given :

In $\Delta ABC,\angle A={60}^{\circ },\angle B={80}^{\circ }$

We know that, sum of the angles of a triangle is ${180}^{\circ }$

$\Rightarrow \angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow \angle C={180}^{\circ }-(\angle A+\angle B)$

$={180}^{\circ }-({60}^{\circ }+{80}^{\circ })$

$={180}^{\circ }-{140}^{\circ }$

$\therefore \angle C={40}^{\circ }$

Bisectors of $\angle B$ and $\angle C$ meet $O$.

In figure,

$BO$ and $CO$ are the bisectors of $\angle B$ and $\angle C$.

$\Rightarrow \angle BCO=\frac{{40}^{\circ }}{2}={20}^{\circ }$

$\angle CBO=\frac{{80}^{\circ }}{2}={40}^{\circ }$

Now, In $\Delta BOC$

$\angle BCO+\angle CBO+\angle BOC={180}^{\circ }$ [Angle sum property of triangle]

$\Rightarrow {20}^{\circ }+{40}^{\circ }+\angle BOC={180}^{\circ }$

$\Rightarrow {60}^{\circ }+\angle BOC={180}^{\circ }$

$\Rightarrow \angle BOC={180}^{\circ }-{60}^{\circ }$

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

Alternate method:

In a $\Delta ABC,$ if the bisectors of $\angle B$ and $\angle C$ meet at $O$, then $\angle BOC={90}^{\circ }+\frac{1}{2}\angle A$

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

$={90}^{\circ }+(\frac{1}{2}\times {60}^{\circ })$

$={90}^{\circ }+{30}^{\circ }$

$={120}^{\circ }$

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