In the following figure, angle X=62°, angle XYZ=54°. If YO and ZO are the bisectors of angle XYZ and angle XZY respectively of triangle XYZ, find angle OZY and angle YOZ.
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Answer:
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 = 32º (OZ is the angle bisector of ∠XZY)
Similarly, ∠OYZ == 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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