Question

In the given figure $AB$ ∥ $CD$ ∥ $EF$, $\angle DEF={45}^{\circ }$ and $AE$ ⊥ $AB$. Also, $\angle BAE={90}^{\circ }$. Then, choose the correct values of $\angle x$, $\angle y$ and $\angle z$.

• A. $\angle x={125}^{\circ }$
  $\angle y={125}^{\circ }$
  $\angle z={55}^{\circ }$
• B. $\angle x={135}^{\circ }$
  $\angle y={135}^{\circ }$
  $\angle z={55}^{\circ }$.
• C. $\angle x={125}^{\circ }$
  $\angle y={125}^{\circ }$
  $\angle z={45}^{\circ }$.
• D. $\angle x={135}^{\circ }$
  $\angle y={135}^{\circ }$
  $\angle z={45}^{\circ }$.

Answer 1

The correct option is D $\angle x={135}^{\circ }$

$\angle y={135}^{\circ }$
$\angle z={45}^{\circ }$.
Since $CD$ ∥ $EF$,
$\angle DEF+\angle y={180}^{\circ }$ (co-interior angles are supplementary)
$\therefore \angle y={180}^{\circ }-{45}^{\circ }={135}^{\circ }$

Since $AB$ ∥ $CD$,
$\angle y=\angle x$ (corresponding angles are equal)
$\therefore \angle x={135}^{\circ }$

Since $AB$ ∥ $EF$,
$\angle BAE+\angle AEF={180}^{\circ }$ (co-interior angles are supplementary)
$\angle AEF={180}^{\circ }-\angle BAE$
$\angle AEF={180}^{\circ }-{90}^{\circ }={90}^{\circ }$
We know that:
$\angle AEF=\angle z+{45}^{\circ }$
$\angle z={90}^{\circ }-{45}^{\circ }$
$\therefore \angle z={45}^{\circ }$