I found this question in an exam question paper.
But I was stumped by it. It seemed that all the answers were correct.
I tried:

Let $\angle ADE=x$ and $\angle EDC=y$.
$\angle DAB=90^{\circ}=\angle DAE+\angle EAB=30^{\circ}+60^{\circ}$.
$\text{ref}\angle AEB=300^{\circ}$.
And we know that $x+y=90^{\circ}$
But now I am stuck. How to proceed?

  • $\begingroup$ Explain the downvote, downvoter. $\endgroup$ Sep 14, 2015 at 4:20
  • $\begingroup$ See the edit, retract the downvote please. $\endgroup$ Sep 14, 2015 at 4:21
  • $\begingroup$ Related: math.stackexchange.com/questions/907837/… $\endgroup$
    – Blue
    Sep 14, 2015 at 4:22
  • 1
    $\begingroup$ Please dont mark duplicate. It is indeed different. $\endgroup$ Sep 14, 2015 at 4:23
  • $\begingroup$ We have $x = \frac{180^\circ-30^\circ}{2} = 75^\circ$. So, $y = 15^\circ$. Then, $\angle DEC = 150^\circ$. $\endgroup$
    – GAVD
    Sep 14, 2015 at 4:30

2 Answers 2


$\triangle AEB$ is equilateral, so $|AE|=|AB|=|AD|$. Hence $\triangle DAE$ is isosceles, so $\angle ADE = \angle AED$, and so $\angle ADE = \frac{1}{2}(\pi-\frac{\pi}{6}) = 5\pi/12$.

The same procedure works on the other side of the square, and hence $\angle DEC = 2\pi-\pi/3-2 \times 5\pi/12 = 5\pi/6 = 150˚ $

  • $\begingroup$ Ohh! I forgot that $BE=BC$!! $\endgroup$ Sep 14, 2015 at 4:32

$DAE$ & $CEB$ are triangles having $BC=BE$ & $AD=AE$ now you can find $x$ ($=75^\circ$) and the others are easy.


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