Question

In the given figure, $\angle X={62}^o,\angle XYZ={54}^o$. In $△XYZ$ If YO and ZO are the bisectors of $\angle XYZ$ and $\angle XZY$ respectively find $\angle OZY$ and $\angle YOZ$.

Answer 1

In $△XYZ$

$\angle XYZ+\angle XZY+\angle YXZ={180}^{\circ }$ (Angle sum property)

${54}^{\circ }+\angle XZY+{62}^{\circ }={180}^{\circ }$

$\angle XZY={180}^{\circ }-{116}^{\circ }={64}^{\circ }$

$\angle OZY=\frac{1}{2}\angle XZY$ (Given $OZ$ is the angle bisector)

$\angle OZY=\frac{1}{2}\times {64}^{\circ }={32}^{\circ }$

$\angle OYZ=\frac{1}{2}\angle XYZ$ (Given $OY$ is then angle bisector)

$\angle OYZ=\frac{1}{2}\times {54}^{\circ }={27}^{\circ }$

In $△OYZ$

$\angle YOZ+\angle OYZ+\angle OZY={180}^{\circ }$

${54}^{\circ }+\angle XZY+{62}^{\circ }={180}^{\circ }$

$\angle YOZ+{27}^{\circ }+{32}^{\circ }={180}^{\circ }$

$\angle YOZ={180}^{\circ }-{59}^{\circ }={121}^{\circ }$