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2. In Fig. 6.40, \angle X = 62°, \angle XYZ = 54°. 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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We have 
\angleX = 62^0\angleXYZ = 54^0
YO and ZO bisects the \angleXYZ and \angleXZY
Now, In \DeltaXYZ, by using angle sum property
\angleXYZ + \angleYZX + \angleZXY = 180^0

So, \angleYZX = 180^0-54^0-62^0
      \angleYZX = 64^0

and, \angleOYZ = 54^0/2 = 27^0 also, \angleOZY = 32^0

Now, in \DeltaOYZ 
\angleY + \angleO + \angleZ = 180^0  [\angleY = 32^0 and \angleZ = 64^0]
So, \angleYOZ = 121^0

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manish painkra

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