Question

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.​

Answer 1

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

Answer 2

$\implies$$\large\sf{\angle{X}+\angle{Y}+\angle{Z}=180°}$

$\implies$$\large\sf{62°+54°+\angle{Z}=180°}$

$\implies$$\large\sf{116°+\angle{Z}=180°}$

$\implies$$\large\sf{\angle{Z}=180°-116°}$

$\implies$$\large\sf{\angle{Z}=64°}$

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$\large\sf{\angle{Z}=64°}$

$\large\sf{\angle{OZY}=\frac{1}{2}\angle{Z}}$

$\large\sf{\angle{Z}=\frac{1}{2}(64°)}$

$\large\sf{\angle{OZY}=32°}$

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$\large\sf{\angle{Y}=54°}$

$\large\sf{\angle{OYZ}=\frac{1}{2}(54°)}$

$\large\sf{\angle{OYZ}=27°}$

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$\implies$$\large\sf{\angle{OZY}+\angle{OYZ}+\angle{YOZ}=180°}$

$\implies$$\large\sf{32°+27°+\angle{YOZ}=180°}$

$\implies$$\large\sf{59°+\angle{YOZ}=180°}$

$\implies$$\large\sf{\angle{YOZ}=180°-59°}$

$\implies$$\large\sf{\angle{YOZ}=121°}$