What is the notion of directed angles in Olympiad Geometry?

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I was reading the book EGMO by Evan Chen when I came across this particular concept of directed angles where a directed angle is defined with the symbol ∡ABC. As far as I could understand, it is defined as an angle mod 180 degrees such that for any angle ABC, ∡ABC is congruent to ∡ABC ± n(180 degrees) which is just like how we define trig functions in a unit circle (Am I correct?).

The well-known theorems have been "re" described following the same understanding. In the book, one example is the Inscribed Angle Theorem: if circle ABC has centre P, then ∡APB = 2∡ACB.

My Question: Does the above theorem precisely say that regardless of whether angle ACB is subtended by major or minor arc, the ∡APB = 2∡ACB ? i.e., neglecting the fact that if the angle is subtended by the major arc, 2∠ACB = 360 - ∠APB (a.k.a reflex ∠APB) ?

I looked at various resources online but I am not clear if I am correct in my understanding (actually I find it hard to digest that these two things mean the same). If somebody can explain it more explicitly, it would be very helpful. By the way, is this concept really essential for Olympiad geometry?

I read other answers but none explicitly explained the answer to the problem I seek.

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Directed angles are defined by two intersecting lines, one of which is designated the initial line and the other of which is designated the terminal line.

With just this information, it makes no sense to talk about subtending major/minor arcs of circles (which is a distinction that comes from the geometry of two rays from the center of a circle)

In terms of directed angles, your equation $$2\angle ACB = 360^\circ - \angle APB \text{ (where this is interpreted as the reflex angle)}$$ would be written $$2\measuredangle ACB = 360^\circ - \measuredangle BPA.$$

Now, since $\measuredangle BPA = -\measuredangle APB$, $$2\measuredangle ACB = 360^\circ + \measuredangle APB$$ and since these are now angles between lines, $360^\circ = 0^\circ$, $$2\measuredangle ACB = \measuredangle APB,$$ so we see the two cases you had before collapse into a single case this context.