Question

In the adjoining figure angle x =62° angle xyz =54° if Yo and zo are the bisectors of anglr xyz respectively of triangle Xyz then find angle ozy and angle yoz​

Answer 1

→ In triangle xyz,

→ Angle x = 62°

→ Angle xyz = 54°

→ Angle x + angle xyz + angle z = 180° [by angle addition property of triangle]

→ 62° + 54° + angle z = 180°

→ 116°+angle z = 180°

→ Angle z = 180-116

→ Angle z = 64°

→ oz is angle bisector of angle xyz

→ Angle ozy = 32° •••••••••• (1) eq

→ Similarly oz is the angle bisector of angle xyz

→ Angle oyz = 27°••••••••••(2) eq

→ In triangle oyz,

→ Angle oyz + ozy + yoz = 180°[ by angle sum property of triangle]

→ 27°+32°+ angle yoz = 180°

→ 59°+angle yoz=180°

→ Angle yoz = 180°+59°

→ Angle yoz = 121°

→ Therefore, angle ozy =32° and angle yoz = 121°

$a\beta hira$

Answer 2

Given:

∠X = 62°

∠XYZ= 54°

To Find:

∠OZY and ∠YOZ

Solution:

given that,

∠X = 62° and ∠XYZ = 54°

∠XYZ + ∠XZY + ∠YXZ = 180°

[angle sum property of triangle]

⟹ 54° + ∠XYZ + 62° = 180°

⟹ ∠XZY + 116° = 180°

⟹ ∠XZY = 180° - 116° = 64°

now, ∠OZY$= \frac{1}{2} \times$ ∠XZY

[ ∵ ZO is bisector of ∠XZY ]

$⟹ \frac{1}{2} \times 64° = 32°$

Similarly,

∠OYZ = $\frac{1}{2} \times 54° = 27°$

Now,in ∆OYZ,we have

∠OYZ + ∠OZY + ∠YOZ = 180°

[angle sum property of triangle]

⟹ 27° + 32° + ∠YOZ = 180°

⟹∠YOZ = 180° - 59° = 121°

Hence,

${\boxed{\tt{\red{∠OZY\:=\:32°}}}}$

${\boxed{\tt{\red{∠YOZ\:=\:121°}}}}$