"World's Hardest Easy Geometry Problem" This question is a "corollary" (if you will) to the World's Hardest Easy Geometry Problem (external website). Formally, this is called Langley's Problem. The objective of that problem was to solve for angle $x^{\circ}$, with the given angles of $10^{\circ}, 70^{\circ}, 60^{\circ}, 20^{\circ}$. Someone presented a solution to that problem. Here's also a rather colorful and interactive solution to a problem like this, but with different angles.
Now, I wanted to generalize this problem, replacing the angles of $10^{\circ}, 70^{\circ}, 60^{\circ}, 20^{\circ}$ with angles of $W^{\circ}, X^{\circ}, Y^{\circ}, Z^{\circ}$, respectively (see below picture). 

How can we derive an analytical expression of angle $x^{\circ}$, in terms of $W^{\circ}, X^{\circ}, Y^{\circ}, Z^{\circ}$?
 A: Well, I'll give a guide to follow, not a final expression.
I called you unknown angle as $\alpha_9$ (in order not to mess, because you have a big $X$ and a small $x$).
Since you know $x$ and $y$ then you know $\alpha_4$.
Since you know $x$, $w$, $z$ and $y$ then you know $\alpha_1$.
Since you know $x$ and $\alpha_5$ then you know $\alpha_3$.
Since you know $w$ and $\alpha_5$ then you know $\alpha_6$.
Then you end up with a system of $4$ equations:
$$
\begin{cases}
\alpha_1+\alpha_2+\alpha_7=\pi \\
\alpha_4+\alpha_8+\alpha_9=\pi \\
\alpha_2+\alpha_3+\alpha_9=\pi\\
\alpha_6+\alpha_7+\alpha_8=\pi
\end{cases} 
$$
with $4$ unknowns: $\alpha_2, \alpha_7, \alpha_8, \alpha_9$.
And $\alpha_9$ is what you are looking for.


A: I have found the equation!
I first tried to insert the latex equation directly here, but that was ugly so I rendered it separately and inserted an image instead:

A: There is in principle no problem in obtaining an expression for the top angle in terms of the bottom ones. Let the unlabelled intersection point in the picture be $I$. Let the bottom side be $1$. We can use the Sine Law to find $AI$ and $BI$. We can also find $AD$, and $BE$, and now we know $ID$ and $IE$, so we can solve for the mystery angle.
This does not result in a nice expression for the mystery angle, but it is an expression. 
A: If one draws a perpendicular line from A and extends line DE to an intersection at, say X, the two resultant triangles allow the equal angles XEA and AEB to be determined at 40 degrees. Angle BED is then 100 degrees and 'x' is 30 degrees.
Thinking outside the triangle(s)?
A: Refering to the Triangle Problem 1 diagram: If z>=w then x=80-w+arctan[(tan(z+10)-tan(w+10))/(tan(10)^2-tan(z+10)*tan(w+10))] degrees. There is no general solution for x in terms of z and w without using trigonometry since some solutions for x require an infinite series of digits to the right of the decimal point.
