$ABC$ is a triangle is which $\angle A = 72^o$, the internal bisectors of angles $B$ and $C$ meet in $O$. Find the magnitude of $\angle BOC$.
Given:
$ABC$ is a triangle is which $\angle A = 72^o$, the internal bisectors of angles $B$ and $C$ meet in $O$.
To do:
We have to find the magnitude of $\angle BOC$.
Solution:
We know that,
Sum of the angles in a triangle is $180^o$.
In $\triangle ABC$,
$\angle B + \angle C = 180^o - 12^o = 108^o$
$OB$ and $OC4 are the bisectors of $\angle B$ and $\angle C$ respectively.
This implies,
$\angle OBC + \angle OCB = \frac{1}{2}(B + C)$
$= \frac{1}{2}(108^o)$
$= 54^o$
In $\triangle OBC$,
$\angle OBC + \angle OCB + \angle BOC = 180^o$
$54^o + \angle BOC = 180^o$
$\angle BOC = 180^o-54^o= 126^o$
The magnitude of $\angle BOC$ is $126^o$.
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