Question

In Fig. 6.40, X = 62°, XYZ = 54°. If YO and ZO are the bisectors of XYZ and
XZY respectively of XYZ, find OZY and YOZ.
​

Answer 1

Given :

• ∠XYZ = 54°
• ∠X = 62°

To Find :

• ∠OZY and ∠YOZ

Solution :

In Δ XYZ :

$\longmapsto\tt{\angle{YXZ}+\angle{XYZ}+\angle{XZY}=180^{\circ}\:(A.S.P)}$

$\longmapsto\tt{62^{\circ}+54^{\circ}+\angle{XZY}=180^{\circ}}$

$\longmapsto\tt{116+\angle{XZY}=180^{\circ}}$

$\longmapsto\tt{\angle{XZY}=180^{\circ}-116^{\circ}}$

$\longmapsto\tt{\angle{XZY}=180^{\circ}-116^{\circ}}$

$\longmapsto\tt\bf{\angle{XZY}=64^{\circ}}$

Also ,

$\longmapsto\tt{\angle{OYZ}=\cancel\dfrac{54}{2}=27^{\circ}}$

$\longmapsto\tt\bf{\angle{OZY}=\cancel\dfrac{64}{2}=32^{\circ}}$

Now ,

In Δ YOZ :

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

$\longmapsto\tt{27^{\circ}+32^{\circ}+\angle{YOZ}=180^{\circ}}$

$\longmapsto\tt{59+\angle{YOZ}=180^{\circ}}$

$\longmapsto\tt{\angle{YOZ}=180^{\circ}-59^{\circ}}$

$\longmapsto\tt\bf{\angle{YOZ}=121^{\circ}}$

Answer 2

Proper Question :

• In the given figure [ as attachment ] ∠62°, ∠XYZ = 54°.If YO and ZO are the bisectors of XYZ and XZY respectively of $\triangle$ XYZ, Find ∠ OZY and ∠YOZ .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given : Angle X = 62⁰ , Angle XYZ = 54⁰ & YO and ZO are the bisectors of XYZ and XZY , respectively .

Exigency To Find : Angle OZY and YOZ .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

• YO is the angle bisector of XYZ :

$\qquad \therefore \sf \angle OYZ = \angle XYO \:=\:\dfrac{1}{2} \:\:of\:\angle XYZ \:\:$

$\qquad :\implies \sf \angle OYZ = \angle XYO \:=\:\dfrac{1}{2} \:\:of\:\angle XYZ \:\:$

⠀⠀⠀⠀⠀Here ,

• $\angle$ XYZ = 54⁰ .

$\qquad :\implies \sf \angle OYZ = \angle XYO \:=\:\dfrac{1}{2} \:\:of\:\angle XYZ \:\:$

$\qquad :\implies \sf \angle OYZ = \angle XYO \:=\:\dfrac{1}{2} \:\:\times\: 54^\circ \:\:$

$\qquad :\implies \sf \angle OYZ = \angle XYO \:=\:\dfrac{1}{\cancel {2}} \:\:\times\: \cancel {54^\circ }\:\:$

$\qquad :\implies \sf \angle OYZ = \angle XYO \:=\: 27^\circ \:\:$

$\qquad \therefore \pmb{\underline{\purple{\:\angle OYZ = \angle XYO \:=\: 27^\circ \:\: }} }\bigstar$

⠀⠀⠀⠀⠀AND ,

• ZO is the angle bisector of XZY :

$\qquad \therefore \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{2} \:\:of\:\angle XZY \:\:$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{2} \:\:of\:\angle XZY\qquad \bigg\lgroup \sf{ \:\:Eq^n \:\:1 \:}\bigg\rgroup \:\:$

Now , In $\triangle$ XYZ :

$\qquad \dashrightarrow \sf \angle YXZ \:+ \:\angle XYZ + \angle XZY =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf \angle YXZ \:+ \:\angle XYZ + \angle XZY =\: 180^0 \:\qquad \bigg\lgroup \sf{ \:\:Angle \:Sum \:Property \:of \:Triangle \:\:\: }\bigg\rgroup\:$

$\qquad \dashrightarrow \sf 62^0 \:+ \:54^0 + \angle XZY =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf 116^0 + \angle XZY =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf \angle XZY =\: 180^0 - 116^0 \:\:$

$\qquad \dashrightarrow \sf \angle XZY =\: 64^0 \:\:$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:in\:Eq^n\: 1 \::}}$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{2} \:\:of\:\angle XZY\qquad \bigg\lgroup \sf{ \:\:Eq^n \:\:1 \:: }\bigg\rgroup \:\:$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{2} \:\:of\:\angle XZY \:\:$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{2} \:\:\times\:64^0 \:\:$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\:\dfrac{1}{\cancel {2}} \:\:\times\:\cancel {64^0} \:\:$

$\qquad :\implies \sf \angle XZO = \angle OZY \:=\: 32^\circ \:\:$

$\qquad \therefore \pmb{\underline{\purple{\:\angle XZO = \angle OZY \:=\: 32^\circ \:\: }} }\bigstar$

Now , In $\triangle$ OYZ :

$\qquad \dashrightarrow \sf \angle OYZ \:+ \:\angle OZY + \angle YOZ =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf \angle OYZ \:+ \:\angle OZY + \angle YOZ =\: 180^0 \:\qquad \bigg\lgroup \sf{ \:\:Angle \:Sum \:Property \:of \:Triangle \:\:\: }\bigg\rgroup\:$

$\qquad \dashrightarrow \sf 27^0 \:+ \:32^0 + \angle YOZ =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf 59^0 + \angle YOZ =\: 180^0 \:\:$

$\qquad \dashrightarrow \sf \angle YOZ =\: 180^0 - 59 ^0\:\:$

$\qquad \dashrightarrow \sf \angle YOZ =\: 121 ^0\:\:$

$\qquad \therefore \pmb{\underline{\purple{\: \angle YOZ \:=\: 121^\circ \:\: }} }\bigstar$

$\qquad \therefore \underline {\sf Hence, \:\: \angle YOZ \:\:and \:\angle OZY \:\:is \:equal \:to \:the \bf \: 32^\circ \:\:\& \:\: 121 ^\circ \: \sf ,respectively. }\:$