Question

The bisectors of $\angle B$ and $\angle C$ of quadrilateral ABCD meet at a point $R$ inside the quadrilateral.If $\angle A={95}^{\circ }$ and $\angle D={111}^0,$ find $\angle BRC$

Answer 1

In quadrilateral $ABCD$,

$\angle A={95}^{\circ },\angle D={111}^{\circ }$

$\angle A+\angle D={111}^{\circ }+{95}^{\circ }={206}^{\circ }$

$\angle B+\angle C={360}^{\circ }-{206}^{\circ }={154}^{\circ }$

In $\Delta BRC$,

$\angle \left(BRC\right)+\angle \left(RBC\right)+\angle \left(BCR\right)={180}^{\circ }$

$\angle \left(BRC\right)+\frac{1}{2}(\angle B+\angle C)={180}^{\circ }$

$\angle \left(BRC\right)+\frac{1}{2}\left({154}^{\circ }\right)={180}^{\circ }$

$\angle \left(BRC\right)={180}^{\circ }-{77}^{\circ }$

$={103}^{\circ }$

$\therefore \angle \left(BRC\right)={103}^{\circ }$