Read too many physics books, and you might forget where "left" is!

I tried to find it out, but neither AI nor the web gave me
the answer. Someone already must have figured that out, but
I was not able to find it!

So, from several notes in the web and my own guesses, I put
together this attempt. Please check it! Is it correct?

My attempt:

Cool down cobalt-60 as much as possible (near absolute zero).

With a strong magnetic field, align the nuclei.

We have two directions:

A: from the north pole of the magnetic field to its south pole, and
B: from the south pole of the magnetic field to its north pole.
(B=-A)

That direction of A and B into which more electrons are emitted
during many decays we call the "south pole" of the nuclei.
(This is our definition of "south pole of the nuclei").

Since all nuclei are aligned, let's focus on one nucleus.

An observer now looks at that nucleus so that its south pole is
at the bottom (direction toward his shoes) for him.

The unit vector from the nucleus to the observer is the x axis.
(This is our definition of that x axis.)

Now assume a small macroscopic object at the place of the
nucleus with the same angular momentum as the spin of the
nucleus.

The macroscopic object rotates a bit. Let the x axis be carried
away (rotated around the object) by this rotation until its
projection onto (scalar product with) the original x axis
(unit vector) vanishes (is zero). (Meaning to rotate it by
90 degrees.)

Then, this rotated x axis now points to the /left/ from the
point of view of the observer.

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