Question

in given figure <x=62°,<xyz=54° in∆xyz if yo and zo are the bisectors of <xyz and<xzy respectively find <ozy and <yoz

Answer 1

In ∆xyz,
x + y + z =180°
=>62+54+z=180°
. : z= 180° - 116° =64°

In ∆yoz,
y/2 + z/2 + o =180°
=>54+64/2+o = 180°
=>118/2 + o =180°
=>59+o=180°
: . o = 180°-59°=121°

hence, angle ozy = 64/2=32°
angle yoz = 121° [: . o =121°]

Answer 2

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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