Question

Question 27

If one of the angle of a triangle is ${110}^{\circ }$, then the angle between the bisectors of the other two angles is

a) ${70}^{\circ }$
b) ${110}^{\circ }$
c) ${35}^{\circ }$
d) ${145}^{\circ }$

Answer 1

In $\Delta ABC,\angle A={110}^{\circ }$

We know, that $\angle A+\angle B+\angle C={180}^{\circ }$ [angle sum property of a triangle]
$\Rightarrow \angle B+\angle C={180}^{\circ }-\angle A\Rightarrow \angle B+\angle C={180}^{\circ }-{110}^{\circ }\Rightarrow \angle B+\angle C={70}^{\circ }\dots \dots \left(i\right)$
$\Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C=\frac{70}{2}={35}^{\circ }$ [$\because$ Eq. (i) is divided by 2]
$\Rightarrow \frac{1}{2}(\angle B+\angle C)={35}^{\circ }$
Now, in $\Delta BOC$
$\angle BOC+\angle OBC+\angle OCB={180}^{\circ }$ [angle sum property of a triangle] .... (ii)
$\Rightarrow \angle BOC+\frac{1}{2}(\angle B+\angle C)={180}^{\circ }$
[$\because$ OB and OC are the bisectors of $\angle B$ and $\angle C$, then $\angle OBC=\frac{1}{2}\angle B$ and $\angle OCB=\frac{1}{2}\angle C$]
$\Rightarrow \angle BOC+{35}^{\circ }={180}^{\circ }\Rightarrow \angle BOC={180}^{\circ }-{35}^{\circ }\Rightarrow \angle BOC={145}^{\circ }$