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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,$32^o+27^o+\angle O=180^o59^o+\angle O=180^o\angle O=180^o-59^o\angle O=121^o$Hence,$\angle OZY=32^o$and$\angle YOZ=27^o\$.

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Updated on: 10-Oct-2022

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