Puzzle time - Find the angle

[jon-nyc]

Definitely there is a solution

[jon-nyc]

No cheating and saying ‘since it’s true for any relative size triangles let me assume they’re the same size and figure out that special case’.

[Klaus]

Oh, I see. The relative size doesn't matter. It's always 120 degrees.

[jon-nyc]

Can you prove it?

[Klaus]

Well, I assume if I set up a formula for the angle given side length a and b for the triangles, then simplify using textbook identities for trigonometric functions, the a's and b's will magically cancel each other all out and I'll just get the 120 degrees.

There should be a tool online that turns geometric constructions into algebraic equations. There's pizza in the oven, so I can't be bothered to do that by hand right now

[Doctor Phibes]

I'm going with 120 degrees.

My reasoning is that Klaus knows about this stuff.

[Doctor Phibes]

Sorry, I should probably have put that in a spoiler

[jon-nyc]

You can do it without complex trig identities. In your head in fact.

[Klaus]

The sum of angles in the upper quadrilateral is 360 degrees. The top one must be 60 degrees. Beta and Gamma add up to 180 degrees. That leaves the desired 120 degrees. The only missing step is why beta and gamma add up to 180.

Something like that?

[jon-nyc]

Not quite right but the solution is that ‘simple’.

[jon-nyc]

My answer:

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[Klaus]

Nice! I'd add a little argument why the two angles you denote by theta are equal.

[jon-nyc]

Ok. Angles 1 and 2 summed together obviously = 180 degrees, but then so do 2 and 3.

If 1+2=180, and 1+3=180, then 2=180-1 and 3=180-1 so 1 and 3 are equal.