In the diagram, $PQ\parallel RS$. Find x in terms of y and z
$x=240{}^\circ -y-z$
$x=180{}^\circ -y-z$
$x=360{}^\circ +y-z$
$x=360{}^\circ -y-z$
$\begin{align} & \text{Construction: Draw line }MN\text{ to pass through O, }\parallel \text{ to QP and SR} \\ & \angle DCO=180{}^\circ -\angle OCP\left| \text{Sum of }\angle \text{s on a straight line} \right. \\ & \angle DCO=180{}^\circ -y \\ & \angle ABO=180{}^\circ -\angle OBR\left| \text{Sum of }\angle \text{s on a straight line} \right. \\ & \angle ABO=180{}^\circ -z \\ & \angle CON=\angle DCO=180{}^\circ -y\left| \text{Alternate angles} \right. \\ & \angle BON=\angle ABO=180{}^\circ -z\left| \text{Alternate angles} \right. \\ & \angle COB=\angle DCO+\angle ABO \\ & \angle COB=180{}^\circ -y+180{}^\circ -z \\ & \angle COB=360{}^\circ -y-z \\ & \angle DOA=\angle COB=360{}^\circ -y-z\left| \text{vertically opposite angles} \right. \\ & \angle DOA=x=360{}^\circ -y-z \\\end{align}$