In the given figure,if $PQ \parallel RS$,$YZ \perp RS$ and $\angle SZX = 143^{\circ}$,then find $\angle PXZ$,$\angle XZY$ and $\angle YXZ$.

(N/A) $1$. Given: $PQ \parallel RS$,$YZ \perp RS$ (i.e.,$\angle YZS = 90^{\circ}$),and $\angle SZX = 143^{\circ}$.
$2$. Since $RS$ is a straight line,$\angle RZX + \angle SZX = 180^{\circ}$ (Linear pair).
$\angle RZX + 143^{\circ} = 180^{\circ} \implies \angle RZX = 37^{\circ}$.
$3$. Since $PQ \parallel RS$,the alternate interior angles are equal. Thus,$\angle PXZ = \angle RZX = 37^{\circ}$.
$4$. Since $YZ \perp RS$,$\angle YZS = 90^{\circ}$. Also,$\angle XZY + \angle YZS = \angle XZS$. Since $\angle XZS$ and $\angle RZX$ are vertically opposite angles,$\angle XZS = \angle RZX = 37^{\circ}$. However,looking at the geometry,$\angle XZY = \angle YZS - \angle XZS = 90^{\circ} - 37^{\circ} = 53^{\circ}$.
$5$. In $\triangle XYZ$,the sum of angles is $180^{\circ}$. $\angle YXZ + \angle XZY + \angle XYZ = 180^{\circ}$. Since $PQ \parallel RS$ and $YZ \perp RS$,$\angle XYZ = 90^{\circ}$.
$\angle YXZ + 53^{\circ} + 90^{\circ} = 180^{\circ} \implies \angle YXZ = 180^{\circ} - 143^{\circ} = 37^{\circ}$.