Question

In the given figures, the directed lines are parallel to each other. Find the unknown angles.

Answer 1

if the lines are parallel
$x+{90}^0={180}^0$ are co-interior angles
$x={180}^0-{90}^0={90}^0$
$\angle 2=x$
$\angle 2={90}^0$
We know that the sum of angles of a triangle
$\angle 1+\angle 2+{30}^0={180}^0$
Substituting the values
$\angle 1+{90}^0+{30}^0={180}^0$
By further calculation
$\angle 1+{120}^0={180}^0$
$\angle 1=180-120={60}^0$
Here $\angle 1=k$ are vertically opposite angles
$k={60}^0$
here $\angle 1=z$ are alternate angles
$z={60}^0$
Here $k+y={180}^0$ are co-interior angles
Substituting the values
${60}^0+y={180}^0$
$y=180-60={120}^0$
Therefore $x={90}^0,y={120}^0,z={60}^0,k={60}^0$.