If you know your basic trig laws, you will notice that LAW OF COSINES is perfect for this problem!
What is law of cosines? In a triangle ABC, with side lengths a, b, c (each opposite to the point), the formula is that:
c^2 = a^2 + b^2 - 2ab*cosC ... where angle C is included between a and b, and c is opposite point C. Without loss of generality, you can "mix and match" a, b, c.
We know DA = 3, and since triangle DEF is equilateral, then summing DA + AF = DF = DE = 5.
Using law of cosines:
AE2=AD2+DE2−2∗AD∗DE∗cosD=9+25−2∗3∗5∗cosD
D = 60 degrees, so cosD = cos(60 degrees) = adjacent/hypotenuse = 1/2
AE2=9+25−2∗3∗5∗12=9+25−15=19
Square rooting both sides, you have AE = √19
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