Question

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. ${50}^0$
• B. ${65}^0$
• C. ${145}^0$
• D. ${155}^0$

Answer 1

The correct option is D $\left({155}^{\circ }\right)$

Let $x^{\circ }$ and $y^{\circ }$ be the bisected angles.

So, in the original triangle $\Delta ABC$, sum of angles $={180}^{\circ }$
Therefore, $2x+2y+130=180$
$\Rightarrow 2(x+y)=50$
$\Rightarrow \angle x+\angle y=25$......(i)

In the smallest triangle $OBC$, we have $\angle x+\angle y+\angle z=180$,
where $z$ is the angle between the bisectors.

Therefore, ${25}^{\circ }+\angle z={180}^{\circ }$ [From equation(i)]
$\Rightarrow \angle z=180-25$
$\therefore \angle z={155}^{\circ }$

Hence, Angle between bisectors of other two angles $={155}^{\circ }$.