Question

In the figure(not drawn to scale), ABCD is a quadrilateral, BED and AEC are straight lines and EB = EC. Find y.

• A. ${66}^o$
• B. ${56}^o$
• C. ${48}^o$
• D. ${86}^o$

Answer 1

The correct option is D ${86}^o$

$ABCD$ is a quadrilateral, $BED$ and $AEC$ are straight lines.

In a $\Delta BCE$, we have

$EB=EC$ .....Given

$\Rightarrow \angle ECB=\angle EBC={62}^{\circ }$ .....Angles opposite to equal sides are also equal

Sum of all angles of a triangle $={180}^{\circ }$

$\Rightarrow \angle ECB+\angle EBC+\angle CEB={180}^{\circ }$

$\Rightarrow {62}^{\circ }+{62}^{\circ }+\angle CEB={180}^{\circ }$

$\Rightarrow \angle CEB={56}^{\circ }$

$\angle CEB=\angle AED={56}^{\circ }$ ....Vertically opposite angles

Now in $\Delta ADE$, we have

Sum of all angles of a triangle $={180}^{\circ }$

$\Rightarrow \angle EDA+\angle AED+\angle y={180}^{\circ }$

$\Rightarrow {38}^{\circ }+{56}^{\circ }+\angle y={180}^{\circ }$

$\Rightarrow \angle y={86}^{\circ }$