Quadrilateral with given angles

my geometric solution: http://www.davdata.nl/math/geopuzzle28.html

arc calculation

arc calculation(2)

Sinx / Sin(x + 64) = Sin38 / Sin84 *Sin22 / Sin48

Multiply both denom and nom by 4Sin82

Sinx /Sin(x + 64) = 4Sin22Sin38Sin82 / (Cos6Sin48Sin82)

(4Sin22Sin38Sin82 = Sin66 REPLACEMENT)

Sinx / Sin(x + 64) = Cos24 / (4Sin48Cos6Sin82)

Sinx / Sin(x + 64) = 1 / (8Sin24Cos6Sin82)

(2Sin24*Cos6 = Sin54 REPLACRMENT)

Sinx / Sin(x + 64) = 1 / (4Sin54*Sin82)

(1/4Sin54 = Sin18 REPLACEMENT)

Sinx / Sin(x + 64) = Sin18 / Sin82

X = 18

Y = 50

A proof by Sine Law goes as follows:

Considering triangles $AED, BED, ABD, ABE$:

begin{align} frac {sin y}{AD} &= frac {sin 48^circ}{ED}\ frac {sin x}{EB} &= frac {sin 38^circ}{ED}\ frac {sin 46^circ}{AD} &= frac {sin 64^circ}{BA}\ frac {sin 22^circ}{EB} &= frac {sin 74^circ}{BA}\ end{align}

Equating $ED$ in the first two equations we have:

$$frac {EB sin 38^circ}{sin x} = frac {AD sin 48^circ}{sin y}$$

$EB$ and $AD$ can be expressed in terms of $BA$:

$$frac {AB sin 22^circ sin 38^circ}{sin 74^circ sin x} = frac {AB sin 46^circ sin48^circ}{sin 64^circ sin y}$$

Noting that $x = 68^circ - y$,

$$frac {sin 22^circ sin 38^circ}{sin 74^circ (sin 68^circ cos y - cos 68^circ sin y)} = frac {sin 22^circ sin 38^circ}{sin 74^circ sin (68^circ - y)} = frac {sin 46^circ sin48^circ}{sin 64^circ sin y}$$

Rearranging:

$$(sin 22^circ sin 38^circ sin 64^circ + sin 46^circ sin 48^circ sin 74^circ cos 68^circ) sin y = sin 46^circ sin 48^circ sin 74^circ sin 68^circ cos y$$

Giving the expression for $tan y$:

$$frac {sin 46^circ sin 48^circ sin 74^circ sin 68^circ} {sin 22^circ sin 38^circ sin 64^circ + sin 46^circ sin 48^circ sin 74^circ cos 68^circ}$$

WolframAlpha says it is $50^circ$. One could probably reduce the expression above, but right now I don't see how.