In the figure,angle X = 62^o and angle XYZ = | vedclass

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(B) In $\Delta XYZ$,the sum of angles is $180^o$.$\angle XYZ + \angle YZX + \angle ZXY = 180^o$Given $\angle XYZ = 54^o$ and $\angle ZXY = 62^o$.$54^o

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$32^o$ and $121^o$

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(B) In $\Delta XYZ$,the sum of angles is $180^o$.$\angle XYZ + \angle YZX + \angle ZXY = 180^o$Given $\angle XYZ = 54^o$ and $\angle ZXY = 62^o$.$54^o + \angle YZX + 62^o = 180^o$$\angle YZX = 180^o - 116^o = 64^o$.Since $YO$ and $ZO$ are bisectors of $\angle XYZ$ and $\angle XZY$ respectively:$\angle OZY = \frac{1}{2} \angle XZY = \frac{1}{2} (64^o) = 32^o$.$\angle OYZ = \frac{1}{2} \angle XYZ = \frac{1}{2} (54^o) = 27^o$.In $\Delta OYZ$,the sum of angles is $180^o$:$\angle YOZ + \angle OYZ + \angle OZY = 180^o$$\angle YOZ + 27^o + 32^o = 180^o$$\angle YOZ = 180^o - 59^o = 121^o$.Thus,$\angle OZY = 32^o$ and $\angle YOZ = 121^o$.

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

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