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?
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Quang Hoang
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Aditya Agarwal
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Explain the downvote, downvoter. – Aditya Agarwal Sep 14 '15 at 04:20
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See the edit, retract the downvote please. – Aditya Agarwal Sep 14 '15 at 04:21
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Related: http://math.stackexchange.com/questions/907837/is-there-a-pure-geometric-solution-to-this-problem – Blue Sep 14 '15 at 04:22
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1Please dont mark duplicate. It is indeed different. – Aditya Agarwal Sep 14 '15 at 04:23
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We have $x = \frac{180^\circ-30^\circ}{2} = 75^\circ$. So, $y = 15^\circ$. Then, $\angle DEC = 150^\circ$. – GAVD Sep 14 '15 at 04:30
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Or, we have $x = 75^\circ$, then $\angle DEC = 360^\circ - 60^\circ - 2\times 75^\circ = 150^\circ$. – GAVD Sep 14 '15 at 04:31
2 Answers
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$\triangle AEB$ 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˚ $
Chappers
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$DAE$ & $CEB$ are triangles having $BC=BE$ & $AD=AE$ now you can find $x$ ($=75^\circ$) and the others are easy.
