XYZ is a triangle YO and ZO are angle bisectors angle YXZ is equal to 62° and angle OYZ is equal to 54° find angle OZY and angle YOZ
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angle ozy=31
angle yoz=95
angle yoz=95
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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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