See the attached figure..
In \(\triangle CDE\), all angles are 60, so \(\angle EDC = 60\)
But \(\angle ADC=90=\angle ADE+\angle EDC=\angle ADE+60....\angle ADE=30\)

Now take triangle ADE, where AD=DE, so ADE is an isosceles \(\triangle\)....\(\angle DAE=\angle AED=x\)
So, x+x+30=180...2x=150...x=75

D

as CDE is equilateral, two conclusions can be made:
(1) ∠EDC = 60 and ∠ADE = 30
(2) Triangle ADE is isosceles triangle where ∠DAE = ∠AED

then ∠DAE = ∠AED = \(\frac{180-30}{2} = 75\) D

Triangle EDC is equilateral, which means that all angles are equal to 60.
∠EDC and ∠ADE are complementary, meaning that
∠EDC=90-∠ADE=90-60=30
∠DAE=∠AED= (180-30)/2=75
Ans=D

I have a query.

Refer the figure below.


In the figure DE = EC = CD = a. EF is perpendicular to AD and EG is perpendicular to DC.
Angle DEF = 60°. EG = DF = \(\frac{\sqrt{3}}{2}*a\). AF = \(a - \frac{\sqrt{3}}{2}*a\).
Since EG is perpendicular to DC, G bisects DC.
Hence DG = FE = \(\frac{a}{2}\)

Now, in triangle AEF, angle AEF = y
Thus, Tan y = AF/EF = \((a - \frac{\sqrt{3}}{2}*a)/\frac{a}{2}\)
Tan y = 1 - \(\frac{\sqrt{3}}{2}\)

Till this point it was clear that I need to know trigonometry to find 'y' but it was difficult to find exact value.
Since n know the angle an be found using trigonometry and it is greater than 60°, chose D(E is not possible).

Is it right to do this way.?
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Can you please explain how did you arrive at the conclusion that Triangle ADE is isosceles triangle?

In triangle EDC, ED = DC = CE. Also, n square ABCD, AB = BC = CD = DA.

From above you can see that DA = EC in triangle.

Hope it's clear.
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