Question

In the adjoining figure X=60o Y=58o . If YO and ZO are bisectors of XYZ and XZY respectively of ∆XYZ, find OZY and YOZ.

Answer 1

Given :

• ∠X = 60°
• ∠Y = 58°

To Find :

• ∠OZY and ∠YOZ .

Solution :

$\longmapsto\tt{\angle{X}+\angle{Y}+\angle{Z}=180^{\circ}\:(A.S.P)}$

$\longmapsto\tt{\angle{60^{\circ}}+\angle{58^{\circ}}+\angle{Z}=180^{\circ}}$

$\longmapsto\tt{118^{\circ}+\angle{Z}=180^{\circ}}$

$\longmapsto\tt{\angle{Z}=180^{\circ}-118^{\circ}}$

$\longmapsto\tt\bf{\angle{Z}=62^{\circ}}$

Now ,

$\longmapsto\tt{\angle{OYZ}=\dfrac{58}{2}=29^{\circ}}$

Similarly ,

$\longmapsto\tt\bf{\angle{OZY}=\dfrac{62}{2}=31^{\circ}}$

In Δ XYZ :

$\longmapsto\tt{\angle{OZY}+\angle{OYZ}+\angle{YOZ}=180^{\circ}\:(A.S.P)}$

$\longmapsto\tt{\angle{31^{\circ}}+\angle{29^{\circ}}+\angle{YOZ}=180^{\circ}}$

$\longmapsto\tt{60^{\circ}+\angle{YOZ}=180^{\circ}}$

$\longmapsto\tt{\angle{YOZ}=180^{\circ}-60^{\circ}}$

$\longmapsto\tt\bf{\angle{YOZ}=120^{\circ}}$

Answer 2

• ∠ozy = 31°
• ∠yoz = 120°

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Explanation :

$\underline{\underline{\sf{\blue{Given\::-}}}}$

• ∆ xyz

In which,

• yo is bisector of ∠xyz
• zo is bisector of ∠xzy
• ∠x = 60°
• ∠y = 58°

$\underline{\underline{\sf{\blue{To\:Find\::-}}}}$

• ∠ozy = ?
• ∠yoz = ?

$\underline{\underline{\sf{\blue{Solution\::-}}}}$

✇ Using angle sum property of triangle on ∆ xyz :-

$\qquad\leadsto\quad\sf \angle{x} + \angle{y} + \angle{z} = 180^{\circ}$

$\qquad\leadsto\quad\sf 60^{\circ} + 58^{\circ} + \angle{z} = 180^{\circ}$

$\qquad\leadsto\quad\sf 118^{\circ} + \angle{z} = 180^{\circ}$

$\qquad\leadsto\quad\sf \angle{z} = 180^{\circ} - 118^{\circ}$

$\qquad\leadsto\quad{\bf{\pink{ \angle{z} = 62^{\circ}}}}$

• As zo is bisector of ∠z

Therefore,

$\qquad\leadsto\quad\sf \angle{ozy} = \dfrac{\angle{z}}{2}$

$\qquad\leadsto\quad\sf \angle{ozy} = \dfrac{62^{\circ}}{2}$

$\qquad\leadsto\quad\sf \angle{ozy} = \dfrac{\cancel{62^{\circ}}}{\cancel{2}}$

$\qquad\leadsto\quad{\bf{\purple{ \angle{ozy} = 31^{\circ}}}}$

Also,

• yo is bisecror of ∠y

Therefore,

$\qquad\leadsto\quad\sf \angle{oyz} = \dfrac{\angle{y}}{2}$

$\qquad\leadsto\quad\sf \angle{oyz} = \dfrac{58^{\circ}}{2}$

$\qquad\leadsto\quad\sf \angle{oyz} = \dfrac{\cancel{58^{\circ}}}{\cancel{2}}$

$\qquad\leadsto\quad{\bf{\green{ \angle{oyz} = 29^{\circ}}}}$

Now,

✇ Using angle sum property of triangle on ∆ oyz :-

$\qquad\leadsto\quad\sf \angle{yoz} + \angle{ozy} + \angle{oyz} = 180^{\circ}$

$\qquad\leadsto\quad\sf \angle{yoz} + 31^{\circ} + 29^{\circ} = 180^{\circ}$

$\qquad\leadsto\quad\sf \angle{yoz} + 60^{\circ} = 180^{\circ}$

$\qquad\leadsto\quad\sf \angle{yoz} = 180^{\circ} - 60^{\circ}$

$\qquad\leadsto\quad{\bf{\red{ \angle{yoz} = 120^{\circ}}}}$

Hence,

• ∠ozy = 31°
• ∠yoz = 120°

$\underline{\underline{\sf{\blue{Note\::-}}}}$

• Diagram is in attachment !

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