Question

In the figure, $\angle X={62}^o,\angle XYZ={54}^o$. 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

Given, $\angle X={62}^{\circ },\angle XYZ={54}^{\circ }$

$YO$ and $ZO$ are the bisectors of $\angle XYZ$ and $\angle XZY$ respectively.

According to the question

$\angle X+\angle XYZ+\angle XZY={180}^{\circ }$ (Sum of the interior angles of the triangle.)

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

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

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

Now, $\angle OZY=\frac{1}{2}\angle XZY$(ZO is the bisector.)

$\Rightarrow \angle OZY={32}^{\circ }$

Also $\angle OYZ=\frac{1}{2}\angle XYZ$ ($YO$ is the bisector.)

$\Rightarrow \angle OYZ=27°$

Now, $\angle OZY+\angle OYZ+\angle O={180}^{\circ }$ (Sum of the interior angles of the triangle.)

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

$\Rightarrow {59}^{\circ }+\angle YOZ={180}^{\circ }$

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