In the diagram below an equilateral triangle CDE has been inscribed in

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.?

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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