4 equations, 4 variables. Why can't I solve it?

Factual Questions
1

I found a math/geometry/trig problem online. You’re given a triangle, and some corner angles, and asked to find “X”.

[Album] Finding X - Album on Imgur

Knowing that all interior angle of a triangle add up to 180° and that all angles in a straight line add up to 180°, I came up with 4 equations for the 4 variables - W, X, Y, and Z.

The equations I got are:
W+Z= 110
X+Y = 100
X+W = 80
Y+Z = 130

I can’t solve for any variable, though. If I simply via substitution, I end up with non helpful solutions like 100 = 100 or 0=0 or X=X

I’m not looking for the answer - I know it. I plugged the angles into AutoCAD and measured it. It’s a whole number integer and it makes sense. What I am looking for is

  1. How do I solve it mathmatically?
  2. Why can’t I solve 4 simultaneous equations with 4 variables?
2

It looks like those equations are not all independent, but one of them can be derived from the others, therefore you have 3 equations, not 4.

You know already that Y+Z=130 from the first 3 equations. So you are attempting to solve something like this:

2y=x
2y=x
2 equations, 2 variables.

3

This seems to be the right answer. I tried solving it and got 0=0. I plugged it into one of the LLMs and it came back with {w, x, y, z} = {30, 50, 50, 80}, which works. So, I tried again with {40, 40, 60, 70} (adding and subtracting 10 from the variables) and it still works. So, the problem is underspecified.

Edit: I don’t know why I quoted you. I thought I was responding to the previous poster.

4

ETA: Never mind. Need to think over it when I’m not on my way out the door.

5

That does make sense. If the equations on dependent on each other, there’s not enough information to solve. FWIW, X=40­°

Is there another independent equation I’m missing?

6

I’m thinking there has to be a geometric fact that gives us one more piece of information like base angles of an isosceles triangle are congruent or the law of sines but I’m not seeing any off-hand.

7

It didn’t help me, but there are 3 different isosceles triangles in the puzzle.

https://imgur.com/a/21IxBpb

There just doesn’t seem to be enough information about that line that ends with the unknown angles.

8

I was looking for a right angle somewhere but I didn’t see any.

9

Is W the little angle or the whole angle? And, the triangle that X&W are in looks like a right triangle, but I haven’t tried proving it.

10

Don’t! The third angle is 100 degrees

11

So it is! Before one of the posters had w+x+y+z = 210 and then deleted it, but it seems right to me. Is that the missing equation?

12

Actually, it’s still underspecified with that equation.

13

How did you get x = w = 40 and why can’t you get y = 70, z = 60 from that?

14

the interior angles of a quadralateral =360° so yes, X+Y+W+Z = 210°

Unfortunately, that gives us no extra information. We know W+Z = 110 and X+Y = 100

So, it’s still the same dependent equation

15

I got X and W by putting the known angles in AutoCAD and measuring it - not solving it mathematically.

Just posted those to point out that X is part of an isosceles triangle. But we have no way of knowing that based on the original problem.

16

There are four right angles right in the middle of the diagram.

17

There are not. 2 of the angles are 100° and 2 are 80°

18

Ah, you are correct.

19

I wish I wasn’t! :grinning_face_with_smiling_eyes:

20

You actually just have two independent equations which is obvious if you put the equations into a matrix form:

\begin{matrix} W & 0 & 0 & Z \\ 0 & X & Y & 0 \\ W & X & 0 & 0 \\ 0 & 0 & Y & Z \\ \end{matrix}

If you add the first row (equation) to the second, and the third to the fourth, you just get W + X + Y + Z for each combination, which is obviously not invertible.

Geometrically, you have no parallel lines or line lengths so you can’t use the law of sines or congruency, and nothing about how this problem is described gives right triangles. However, if you notice that you have two angles of a triangle (50º and 80º) so you know the third angle is 50º, so you have an isosceles triangle (two sides are the same length, and bisecting between them gives you two mirrored right triangles…the figure is misleading to say the least). I haven’t worked through the problem but I believe that you can use that to define the relative lengths of the sides and from there calculate W, X, Y, and Z using algebra and trigonometric relationships of the bisected right triangle.

Stranger