Question

in given figure angle x = 62 degree, angle XYZ is=54 degree. if YO and ZO are the bisectors of angle XZY and angle XYZ respectively of triangle XYZ. find angle OZY and YOZ.

Answer 1

Hello mate ☺

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Solution:

In ∆XYZ, we have

∠XYZ+∠XZY+∠X=180°    (Sum of three angles of a triangle =180°)

⇒540+∠XZY+62°=180°

⇒∠XZY=180°−54°−62°=64°

It is given that OY and OZ are bisectors of ∠XYZ and ∠XZY respectively

Therefore, ∠OZY=1/2(∠XZY)

=1/2(64°)

=32°

Similarly, ∠OYZ=1/2(∠XYZ)

=1/2(54°)

=27

In ∆OYZ, we have

∠OYZ+∠OZY+∠YOZ=180°  (Sum of three angles of a triangle =180°)

⇒27°+32°+∠YOZ=18°

⇒∠YOZ=180°−27°−32°=121°

Therefore, ∠OZY=32° and ∠YOZ=121°

I hope, this will help you.☺

Thank you______❤

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Answer 2

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

Step-by-step explanation:

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