>>13668124This. Everyone else itt is a brainlet. I solved it differently, but that just goes to show there are multiple ways of arriving at the same answer.
Here's how I did it. I labeled AE as x, DE as y, and DF as z to start. Clearly, angle EDF is pi/6 and angle ADE is right. Exploiting this latter fact, we can already eliminate a variable:
x^2=y^2+1^2
x=sqrt(y^2+1)
Then I just used the law of cosines twice - once for the triangle DEF knowing the angle EDF is pi/6 and again for the triangle ADF knowing the angle ADF is 2pi/3:
(1) 1^2=y^2+z^2-2*y*z*cos(pi/6)
(2) (1+sqrt(y^2+1))^2=1^2+z^2-2*z*cos(2pi/3)
There are two equations and two unknowns. Solving (1) gives:
z=(sqrt(3)*y+sqrt(4-y^2))/2
Plugging this into (2) gives:
(1+sqrt(y^2+1))^2=1+((sqrt(3)*y+sqrt(4-y^2))/2)^2+(sqrt(3)*y+sqrt(4-y^2))/2
At this point I just threw this into Wolfram Alpha, which returned:
y=(1-2^(1/3)+2^(2/3))/sqrt(3)
But squaring this and cancelling some terms gives:
y^2 = 2^(2/3)-1
Plugging this into the expression for x above gives:
x=2^(1/3)
Incidentally, even though the problem doesn't ask:
z=2^(2/3)=x^2
