Question

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

Answer 1

As the sum of all interior angles of a triangle is ${180}^{\circ }$,

Now, In $\Delta XYZ$,

$\angle YXZ+\angle XYZ+\angle XZY={180}^{\circ }$

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

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

$\angle XZY={64}^{\circ }$

$\Rightarrow \angle OZY=\frac{64}{2}={32}^{\circ }$ [Since $OZ$ is the angle bisector of $\angle XZY$

Similarly, $\angle OYZ=={27}^{\circ }$

Using angle sum property for $\Delta OYZ$, we obtain

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

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

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

$\angle YOZ={121}^{\circ }$