Question

In the given figure, 2X = 72°, ZXZY = 46°. If YO and ZO are bisectors of ZXYZ and ZXZY respectively of AXYZ, find LOYZ and LYOZ. what will be the answer?​

Answer 1

Answer:

Solution-

Given that,

In ∆ XYZ

∠YXZ = 72°

∠XZY = 46°

We know, Sum of all interior angles of a triangle is supplementary.

⇛ ∠XYZ + ∠XZY + ∠YXZ = 180°

∠XYZ +46° + 72° = 180°

∠XYZ + 118° = 180°

∠XYZ = 180° - 118°

⇛ ∠XYZ = 62°

Now, further given that

OY bisects ∠XYZ

⇛∠XYZ = 2∠OYZ

⇛62° = 2∠OYZ

⇛ ∠OYZ = 31°

Also, given that

OZ bisects ∠XZY

⇛∠XZY = 2∠OZY

⇛ 46° = 2∠OZY

⇛∠OZY = 23°

Now, In ∆ OYZ

We know, Sum of all interior angles of a triangle is supplementary.

∠YOZ + ∠OYZ + ∠OZY = 180°

On substituting the values, we get

∠YOZ + 31° + 23° = 180°

∠YOZ + 54° = 180°

∠YOZ = 180° - 54°

⇛∠YOZ = 126°.

Hence,

⇛{ ∠XYZ =62°

{∠ YOZ = 126

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Answer 2

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