Ray YM is the bisector of angle XYZ and ray Y | vedclass

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(D) Let $\angle XYZ = 2	heta$. Since $YM$ is the bisector of $\angle XYZ$,we have $\angle XYM = \angle MYZ = 	heta$.Since $YN$ is the bisector of $\

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) Let $\angle XYZ = 2	heta$. Since $YM$ is the bisector of $\angle XYZ$,we have $\angle XYM = \angle MYZ = 	heta$.Since $YN$ is the bisector of $\angle MYZ$,we have $\angle MYN = \angle NYZ = \frac{	heta}{2}$.Given that $\angle XYN = 45^{\circ}$,we can express this as $\angle XYM + \angle MYN = 45^{\circ}$.Substituting the values,we get $	heta + \frac{	heta}{2} = 45^{\circ}$.This simplifies to $\frac{3	heta}{2} = 45^{\circ}$,which gives $3	heta = 90^{\circ}$,so $	heta = 30^{\circ}$.Therefore,$\angle XYZ = 2	heta = 2(30^{\circ}) = 60^{\circ}$.

Ray $YM$ is the bisector of $\angle XYZ$ and ray $YN$ is the bisector of $\angle MYZ$. If $\angle XYN = 45^{\circ}$,then find $\angle XYZ$. (in $^{\circ}$)

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