In Fig. 6.40, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and XZY respectively of Δ XYZ, find OZY and YOZ.

It is given in the question that:

X = 62o, XYZ = 54o


YO and ZO are the bisectors of XYZ and XZY respectively.


Now, according to the question,


X + XYZ + XZY = 180o (Sum of the interior angles of triangle)


62o + 54o + XZY = 180o


116o + XZY = 180o


XZY = 64o


Now,


OZY =


OZY = 32o


And,


OYZ = XYZ (YO is the bisector)


OYZ = 27o


Now,


OZY + OYZ + O = 180o (Sum of the interior angles of the triangle)


32o + 27o + O = 180o


59o + O = 180o


O = 121o


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