In Delta XYZ,XY = XZ and angle X = 80^{circ}, | vedclass

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(A) Given that in $\Delta XYZ$,$XY = XZ$. Since two sides are equal,$\Delta XYZ$ is an isosceles triangle. Therefore,the angles opposite to the equal

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(A) Given that in $\Delta XYZ$,$XY = XZ$. Since two sides are equal,$\Delta XYZ$ is an isosceles triangle. Therefore,the angles opposite to the equal sides are equal,which means $\angle Y = \angle Z$. We know that the sum of all interior angles in a triangle is $180^{\circ}$. So,$\angle X + \angle Y + \angle Z = 180^{\circ}$. Substituting the given values: $80^{\circ} + \angle Y + \angle Y = 180^{\circ}$. $80^{\circ} + 2\angle Y = 180^{\circ}$. $2\angle Y = 180^{\circ} - 80^{\circ} = 100^{\circ}$. $\angle Y = 100^{\circ} / 2 = 50^{\circ}$.

In $\Delta XYZ$,$XY = XZ$ and $\angle X = 80^{\circ}$,then $\angle Y = \ldots$ (in $^{\circ}$)

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