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In the given figure, X = 62º, XYZ = 54º. If YO and ZO are the bisectors of XYZ and XZY respectively of ΔXYZ, find OZY and YOZ.

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As the sum of all interior angles of a triangle is 180º, therefore, for ΔXYZ,

X + XYZ + XZY = 180º


62º + 54º + XZY = 180º


XZY = 180º − 116º

XZY = 64º

OZY = 64/2 = 32º (OZ is the angle bisector of XZY)


Similarly, OYZ = 54/2 = 27º


Using angle sum property for ΔOYZ, we obtain 

OYZ+ YOZ + OZY = 180º 

27º + YOZ + 32º = 180º


YOZ = 180º − 59º

YOZ = 121º

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