Question

In $△ABC,AB\&CD$ are internal angle bisectors. Given that $\angle A={60}^o\&\angle C={40}^o$. Find the angle subtended by $AC$ on incentre of the triangle.

• A. ${130}^o$
• B. ${150}^o$
• C. ${120}^o$
• D. ${125}^o$

Answer 1

The correct option is A ${130}^o$

$\text{Incentre}$ is the point where all the three interior angle bisectors of a triangle meet.


Here, $O$ is the point where all three angle bisectors, $AE$, $BF$ and $CD$ meet.

Hence, $O$ is the incentre of the triangle $△ABC.$

$\angle OAC=\frac{1}{2}\angle A=\frac{1}{2}\times {60}^o={30}^o$

$\angle OCA=\frac{1}{2}\angle C=\frac{1}{2}\times {40}^o={20}^o$

$\angle OAC+\angle OCA+\angle AOC={180}^o\Rightarrow {30}^o+{20}^o+\angle AOC={180}^o\Rightarrow \angle AOC={180}^o-{50}^o\Rightarrow \angle AOC={130}^o$

So, option(a) is correct.