I was looking for that. Like in Elements where Euclid used a series of steps to show AB was congruent to EF. Didn’t see any.
Yeah, there are no simple congruencies in the way the problem is stated but if you split the triangle into two right triangles and so some algebra with the angles and relationships of the sides I think you can formulate a set of four independent equations that can be solved for W, Z, Y, and Z.
Stranger
I haven’t read any posts past the original.
The solution is
W = 40
X = 40
Y = 60
Z = 70
With these sort of problems, I go straight to vectors. At least for me, it’s easier to construct where things are. I start with defining points.
Point A is the lower left vertex.
Point B is the upper middle vertex.
Point C is the lower right vertex.
Point D is the confluence of angles X and Y.
Point E is the confluence of angles W and Z.
Next, I define some lengths.
length(AB) is 1. This is choosing arbitrary length scale.
length(AC) = c.
length(AD) = d.
length(AE) = e.
length(BC) = f.
length(BD) = g.
length(BE) = h.
Now we can construct some useful vectors.
vector(A->B) = [cos(80), sin(80)].
vector(A->C) = c*[1, 0].
vector(A->D) = d*[cos(30), sin(30)].
vector(A->E) = e*[1, 0].
vector(B->C) = f*[cos(50), -sin(50)].
vector(B->D) = g*[cos(50), -sin(50)].
vector(B->E) = h*[cos(70), -sin(70)].
The vectors looping around each of the triangles ABC, ADE, and BDE must sum to zero.
[0,0] = vector(A->B) + vector(B->C) + vector(C->A)
[0,0] = vector(A->D) + vector(D->E) + vector(E->A)
[0,0] = vector(B->D) + vector(B->E) + vector(B->A)
This gives us six equations with six unknowns, which can be solved with linear algebra.
length(AB) = 1.
length(AC) = cos(80)+sin(80)/tan(50). Value is 1.
length(AD) = (cos(80)/cos(50)+sin(80)/sin(50)) / (cos(30)/cos(50)+sin(30)/sin(50)). Value is about 0.7779.
length(AE) = cos(80)+sin(80)/tan(70). Value is about 0.5321.
length(BC) = sin(80)/sin(50). Value is about 1.2856.
length(BD) = length(AD)*cos(30)/cos(50)-cos(80)/cos(50). Value is about 0.7779.
length(BE) = sin(80)/sin(70). Value is about 1.0480.
So now we have the locations of all five points. We can use dot products to extract the angle-cosines we’re interested in.
vector(E->B) dot vector(E->D) = cos(W)*length(EB)*length(ED).
vector(D->A) dot vector(D->E) = cos(X)*length(DA)*length(DE).
vector(D->E) dot vector(D->C) = cos(Y)*length(DE)*length(DC).
vector(E->D) dot vector(E->C) = cos(Z)*length(ED)*length(EC).
And those cosines give the angles at the top of the post.
A solution seems to be in calculating length ratios. Though you don’t know lengths, you can assume a leg leg length and calculate leg lengths based on that. This will quickly allow you to get the leg lengths from the bottom right corner to point XY, and also point WZ. Once those leg lengths are know that triangle is defined (along with the angle) and Y and Z are thus now known, which unravels the equation.
That works - but it seems like the long way around.
Also, hi Pleonast! I haven’t been on the 'Dope much lately - but yours is a name I recognize. Probably from some Mafia game.
Yup! That way works.
If I stick to Pleonast’s naming scheme - I set length AB to 10. That makes AC 10 as well.
We now have side/angle/side so we can get lengths BD and BC so we can calculate length DC
Do the same to get length EC.
With those two lengths - we can calculate angles Z and Y .
Once you have Y, X is easy.
Not simultaneous equations at all. Just trigonometry.
vote brewha for using trig.
I find working with vectors and linear equations more straightforward than trigonometry. But good math will lead to the same answer, so whatever works best for the person doing it is great.
By the way, did you actually work the numbers to get values for W, X, Y, Z? Or just see that it could be done?
Boooo! Anti trig is something only a wolf would be! 
Yes, I ended up with 59.977° for Y. None of the other side lengths ended up nice numbers, so there were some rounding errors baked in. So, if Y is actually 60°, then X is 40°.
Without actually going through all of the steps, I would arbitrarily assign some side the length 1 (probably the left side, just because it’s sort of symmetric about that, and that side is relatively uncluttered), then use the Law of Sines to get the lengths of the lines from the left and top vertices to the center crossing point, then use Law of Sines again to get the other two line segments at the center crossing point, and then use the Law of Cosines and the Law of Sines to get angles X and W from that small triangle in the center.
Elimination yields three equations:
X=80-W
Y=20+W
Z=110-W
and W arbitrary (although between 0 and 80 to keep all angles positive).
Sure, but because all of the angles in the triangle are specified, it is not really that arbitrary. Otherwise you could indeed try to set W close to 0 and fulfill all of the constraints.
Those aren’t actually totally independent. You can solve one of them for W and substitute it into another so that you just have one equation in terms of X and Y, and the other in terms of W and Z, so there are really only two independent equations, e.g.:
W = Y -20
→ X = 80 - (Y-20)
→ X + Y = 100
Stranger
They look like three equations to me. The rank of the matrix is, in fact 3.
The quantified angles only provide two degrees of freedom (there are four listed angles but two are compliment pairs), so you can only formulate two independent degrees of freedom.
Stranger
If there are only two independent equations, what are they?
Once combination:
X + Y = 100
W + Z = 110
Stranger
Sorry @Stranger_On_A_Train, @Hari_Seldon is right, the matrix has rank 3, not 2. Another equation, independent of the two you wrote, is Z - X = 30. Of course, there are lots of others you could write down as the third equation, but the point is that you can write down three independent equations.
What is going on here, though? Most of these sorts of problems can be solved with basic rules about the sum of angles in a triangle, complementary angles, opposing angles, etc. Here, that doesn’t seem to be the case, even though it does have a clear solution.
The solutions presented so far require fixing a side length and using trig to solve. Even though the problem is independent of scale.
So what’s special about this problem? Are there more subtle angle identities that we’re ignoring? Can problems like this be placed into two different classes, one that’s solvable without referencing side lengths and one that isn’t?
Or is it the typical “found on the internet” math teaser that’s been re-transcribed plagiarized so many times that someone unwittingly changed something and now it’s a broken problem, not a good problem?
There’s nothing wrong with the problem per-se. It has a single, perfectly valid solution. You might say the problem was broken if it were given to elementary school students who had just learned basic rules about the angles in a triangle, etc., but that’s not the case here.
It’s pretty straightforward to solve by fixing one side as length 1 and then using the law of sines, law of cosines, basic trig, etc. to work out the rest. I just find it strange that that’s necessary.
Now that I think about it, this implies that even if you’ve been given the solution (and can thus fill out all remaining angles), simply verifying the various triangle identities isn’t enough to prove that you have a rigid object. Some other constraint is needed on top of that.