Question

ABC is a triangle in which $\angle A={72}^{\circ }$, the internal bisectors of angles B and C meet in O. Find the magnitude of $\angle BOC$.

Answer 1

In $\Delta ABC,\angle A={72}^{\circ }$ and bisectors of $\angle B$ and $\angle$ C meet at O.

Now $\angle B+\angle C={180}^{\circ }-{72}^{\circ }={108}^{\circ }$

$\because$ OB and OC are the bisectors of $\angle B$ and $\angle C$respectively

$\therefore \angle OBC+\angle OCB=\frac{1}{2}(\angle B+\angle C)$
=$\frac{1}{2}\times {108}^{\circ }={54}^{\circ }$
But in $\Delta OBC,$
$\therefore \angle OBC+\angle OCB+\angle BOC={180}^{\circ }$
$\Rightarrow {54}^{\circ }+\angle BOC={180}^{\circ }$
$\therefore \angle BOC={180}^{\circ }-{54}^{\circ }={126}^{\circ }$