Question

In the given figure AB is parallel to ED. If $\angle ABC={160}^{\circ }$ and $\angle EDC={130}^{\circ }$

then find the value of x and y.

• A. $x={90}^{\circ }andy={270}^{\circ }$
• B. $x={80}^{\circ }andy={280}^{\circ }$
• C. $x={70}^{\circ }andy={290}^{\circ }$
• D. $x={60}^{\circ }andy={300}^{\circ }$

Answer 1

The correct option is C

$x={70}^{\circ }andy={290}^{\circ }$

Construction: Draw a line FG through C such that $FG\parallel AB\parallel ED$ as shown.

$\Rightarrow \angle ABC+\angle BCF={180}^{\circ }(Co-interiorangles)$

$\Rightarrow \angle BCF={180}^{\circ }-\angle ABC$

$\Rightarrow \angle BCF={180}^{\circ }-{160}^{\circ }$

$\Rightarrow \angle BCF={20}^{\circ }$

$\Rightarrow \angle FCD+\angle CDE={180}^{\circ }(Co-interiorangles)$

$\Rightarrow \angle FCD={180}^{\circ }-\angle CDE$

$\Rightarrow \angle FCD={180}^{\circ }-{130}^{\circ }$

$\Rightarrow \angle FCD={50}^{\circ }$

$x=\angle BCF+\angle FCD$

$x={20}^{\circ }+{50}^{\circ }={70}^{\circ }$

$y={360}^{\circ }-x\left(Angleaboutapoint\right)$

$y={360}^{\circ }-{70}^{\circ }={290}^{\circ }$