Question

Question 5
If one of the angles of a triangle is ${130}^{\circ }$, then the angle between the bisectors of the other two angles can be:

A) ${50}^{\circ }$
B) ${65}^{\circ }$
C) ${145}^{\circ }$
D) ${155}^{\circ }$

Answer 1

The answer is D.
Let angles of a triangle be $\angle A,\angle Band\angle C$.

$In\Delta ABC,$
$\angle A+\angle B+\angle C={180}^{\circ }$
[Sum of all interior angles of a triangle is ${180}^{\circ }$]
$\Rightarrow \frac{1}{2}\angle A+\frac{1}{2}\angle B+\frac{1}{2}\angle C=\frac{{180}^{\circ }}{2}={90}^{\circ }$
[dividing both sides by 2]
$\Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C={90}^{\circ }-\frac{1}{2}\angle A$
$\Rightarrow [\because in\Delta OBC,\angle OBC+\angle BCO+\angle COB={180}^{\circ }]$
$\left[\begin{array}{c}Since,\frac{\angle B}{2}+\frac{\angle C}{2}+\angle BOC={180}^{\circ }asBOandOCaretheangle\\ bisectorsof\angle ABCand\angle BCArespectively\end{array}\right]$
$\Rightarrow {180}^{\circ }-\angle BOC={90}^{\circ }-\frac{1}{2}\angle A$
$\therefore \angle BOC={180}^{\circ }-{90}^{\circ }+\frac{1}{2}\angle A={90}^{\circ }+\frac{1}{2}\angle A$
$={90}^{\circ }+\frac{1}{2}\times {130}^{\circ }={90}^{\circ }+{65}^{\circ }[\because \angle A={130}^{\circ }(given\left)\right]$
$={155}^{\circ }$
Hence, the required angle is ${155}^{\circ }$.