Drag point A.
But what is the mistake?
Let me walk through the problem. Red text indicates a demonstrable error. What are the conditions for displaying red? From left to right...

∡ θ: By construction, θ cannot go negative. The red channel of the text block:
If[sgn(θ)≥0, 0, .5]

tan θ: In the worksheet, the variable tanθ = sθ/cθ. Using a binary sign
Hence we must have
The red channel reads
If[sgn(s_θ/c_θ) =σ1 σ2, 0, .5]

Can you guess what the problem is? Look at the last red text:

Significant figures.
? This means nothing at the computer. It says "the error in elementary arithmetic operations can be unbounded". Now look below, in the cosine row. -4.4409E -16? Asked for more Significant Figures, GGB will happily fill them in. But the fact is, we are out of precision. These aren't trailing bits which are still good. The bits are junk.
But don't take my word for it. Drag point A. What happens?

This is a mistake in GGB design. The crux is this. Subtractive cancellation can magnify error. Depending on the structure of the problem, it may fall to the user to deal with this in the most effective way, but that's assuming the code implementation is numerically stable. Is it? At present GGB's approach to the problem is perfectly backward:
When the result of -- say -- an inner product is the final value of a computation, trailing bits are politely discarded. Query the number and it behaves (correctly) as having reduced precision. Its behavior is stable. But if that result is an intermediate number in a calculation, (say, it is then used as a divisor), the trailing bits are resurrected.
Now, if the trailing bits met the definition of Significant Digit (they are correctly rounded), then we could at least argue that, in a final calculation, the whole mess of bits are significant. In subsequent calcluations, this notion of "significant digit" breaks down, and is discarded in numerical analysis.
Here, we are in a pickle : the bits GGB passes on to subsequent calculations are not significant. The resulting error cannot be bounded. For nothing prevents all of the nonzero bits of cθ being bad.
Drag point A again.
Here, I will solve the problem with a more careful sign test. I will check the signs of cθ and sθ, and then tell GGB what sign the resulting division is to have.