Question

From the figure given, find the sum of $x,y$ and $z$.

Answer 1

From the figure $BE$ is parallel to $DC$
$\angle EBC+\angle z=(180{)}^{\circ }$ (Co interior angles)

${70}^{\circ }+\angle z=(180{)}^{\circ }$
$\angle z=(180-70{)}^{\circ }={110}^{\circ }$
$\angle ABE=\angle z={110}^{\circ }$ (corresponding angles)
$\angle x={150}^{\circ }-\angle ABE={150}^{\circ }-\angle z={150}^{\circ }-{110}^{\circ }={40}^{\circ }$

Since $AE\parallel BD$, $y=x$ as they are alternate angles.
$\angle y={40}^{\circ }$
The required sum $=x+y+z={40}^{\circ }+{40}^{\circ }+{110}^{\circ }={190}^{\circ }$