Question

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

Answer 1

( Given) In $△xyz$,

$\angle x={62}^o$

$\angle xyz={54}^o$

$OY$ is bisector of $\angle xyz$

$OZ$ is bisector of $\angle xzy$

To find $\to \angle ozy=?\angle \left(yoz\right)=?$

$\Rightarrow$ As we know, sum of angles of $△={180}^o$

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

$62+54+\angle XZY=180$

$\angle XZY=180-116$

$\angle XZY={64}^o$

As $OZ$ is bisector of $\angle xzy$

$\therefore \angle OZY=\frac{\angle XZY}{2}=\frac{{64}^o}{2}$

$\angle OZY={32}^o$

$\Rightarrow$ Now, In $△OZY,\angle YOZ+\angle OYZ+\angle YZO={180}^o$

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

$\angle YOZ=180-59$

$\angle YOZ={21}^o$