Question

In the figure angle X=72° , angle XYZ=46° . If YO and ZO are bisectors of angle xyz and angle xzy respectively of triangle XYZ , find angle OYZ and angle YOZ​

Answer 1

Answer:

This is your answer in that picture

Answer 2

Given :- In ∆XYZ, ∠X = 72° , ∠XZY = 46° . YO and ZO are angle bisectors of ∠XYZ and ∠XZY .

To Find :- ∠OYZ and ∠YOZ = ?

Concept used :-

• By angle sum property of traingles, sum of all three angles of a triangle is equal to 180° .
• Angles bisector of an angle divides the angle in two equal parts .

Solution :-

In ∆XYZ we have,

→ ∠YXZ = 72° { given }

→ ∠XZY = 46° { given }

So,

→ ∠YXZ + ∠XZY + ∠XYZ = 180° { By angle sum property }

then,

→ 72° + 46° + ∠XYZ = 180°

→ 118° + ∠XYZ = 180°

→ ∠XYZ = 180° - 118°

→ ∠XYZ = 62°

therefore,

→ ∠XYO = ∠OYZ = (1/2)∠XYZ { Since YO is angle bisector of ∠XYZ }

→ ∠OYZ = (1/2) × 62°

→ ∠OYZ = 31° (Ans.)

Now, in ∆OYZ we have,

→ ∠OYZ = 31° { solved above }

→ ∠OZY = (1/2)∠XZY = (1/2) × 46° = 23° { Since ZO is angle bisector of ∠XZY }

So,

→ ∠OYZ + ∠OZY + ∠YOZ = 180° { By angle sum property }

→ 31° + 23° + ∠YOZ = 180°

→ 54° + ∠YOZ = 180°

→ ∠YOZ = 180° - 54°

→ ∠YOZ = 126° (Ans.)

Hence, ∠OYZ is equal to 31° and ∠YOZ is equal to 126° .

Shortcut :-

→ ∠YOZ = 90° + (1/2)∠x

→ ∠YOZ = 90° + (1/2) × 72°

→ ∠YOZ = 90° + 36°

→ ∠YOZ = 126° (Ans.)

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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