Question

In the figure (not drawn to scale), $ABCD$ is a square, $ADE$ is equilateral triangle and $BFE$ is a straight line, then find $y$.

• A. ${90}^o$
• B. ${45}^o$
• C. ${75}^o$
• D. ${15}^o$

Answer 1

The correct option is C ${75}^o$

In $△AEB$, we have

$\angle A=\angle DAE+\angle BAD$
$\Rightarrow \angle A={60}^o+{90}^o={150}^o$
And $AE=EB$
$\Rightarrow \angle ABE=\angle AEB$ ....(Angles opposite to equal sides are equal)
Now, $\angle A+\angle ABE+\angle AEB={180}^o$ ....(Angle sum property)
$\Rightarrow 2\angle AEB={180}^o-{150}^o={30}^o\Rightarrow \angle AEB={15}^o$
Now, $\angle E={60}^o$
$\Rightarrow \angle DEF={60}^o-{15}^o={45}^o$
In $△EFD$, we have

$\angle DEF+\angle EDF+\angle EFD={180}^o$
$\Rightarrow {45}^o+{60}^o+y={180}^o$
$\Rightarrow y={180}^o-({45}^o+{60}^o)={75}^o$