In the diagram below an equilateral triangle, CDE has been inscribed inside a square ABCD. What is the value of the angle marked x°?

(A) 15

(B) 30

(C) 45

(D) 75

(E) it can't be determined from the facts given

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
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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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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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