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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} $.
"
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 $\angle YOZ=27^o$.
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