In Fig. $ 6.40, \angle \mathrm{X}=62^{\circ}, \angle \mathrm{XYZ}=54^{\circ} $. If $ \mathrm{YO} $ and $ Z \mathrm{O} $ are the bisectors of $ \angle \mathrm{XYZ} $ and $ \angle \mathrm{XZY} $ respectively of $ \triangle \mathrm{XYZ} $ find $ \angle \mathrm{OZY} $ and $ \angle \mathrm{YOZ} $.
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Given:

$\angle X=62^o$, \angle XYZ=54^o$.

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

To do:

We have to find $\angle OZY$ and $\angle YOZ$.

Solution:

We know that the sum of the interior angles of the triangle are always $180^o$.

This implies,

$\angle X+\angle XYZ+\angle XZY=180^o$

By substituting the values of $X$ and $Y$ we get,

$62^o+54^o+\angle XZY=180^O$

This implies,

$116^o+\angle XZY=180^o$

$\angle XZY=180^o-116^o$

$\angle XZY=64^o$

We know that,

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

Therefore we get,

$\angle OYZ=\frac{1}{2}\angle XZY$

$\angle OYZ=\frac{54^o}{2}$

$\angle OYZ=27^o$

In the similar way we get,

$\angle OZY= \frac{1}{2}\angle XZY$

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

$\angle OZY=32^o$

Since we know that sum of the interior angles of the triangle  are always $180^o$, we get,

$\angle OZY+\angle OYZ+\angle O=180^o$

By substituting the values we get,

$32^o+27^o+\angle O=180^o$

$59^o+\angle O=180^o$

$\angle O=180^o-59^o$

$\angle O=121^o$

Hence, $\angle OZY=32^o$ and $\angle YOZ=27^o$.