Question

Question 2
In the given figure, $\angle X={62}^{\circ }\angle XYZ={54}^{\circ }$. If YO and ZO are the bisector of $\angle XYZ$ and $\angle XZY$ respectively of $\Delta XYZ,$ find $\angle OZY$ and $\angle YOZ.$

Answer 1

Sum of all interior angles of a triangle is ${180}^{\circ }$.
Therefore, in $\Delta XYZ,$
$\angle X+\angle XYZ+\angle XZY={180}^{\circ }$
${62}^{\circ }+{54}^{\circ }+\angle XZY={180}^{\circ }$
$\angle XZY={180}^{\circ }-{116}^{\circ }$
$\angle XZY={64}^{\circ }$
$\angle OZY=\frac{64}{2}={32}^{\circ }$ (OZ is the angle bisector of $\angle XZY$)
Similarly, $\angle OYZ=\frac{54}{2}={27}^{\circ }$
Using angle sum property for $\Delta OYZ,$ we obtain,
$\angle OYZ+\angle YOZ+\angle OZY={180}^{\circ }$
${27}^{\circ }+\angle YOZ+{32}^{\circ }={180}^{\circ }$
$\angle YOZ={180}^{\circ }-{59}^{\circ }$
$\angle YOZ={121}^{\circ }$