About the definition of angle, from Geometric Algebra

In GA, bivectors represent  2-dimensional directions. Any simple bivector can be written as
$$\theta e_1e_2$$
with $\theta \in \mathbb{R}$ and $\{e_i\}$ unitary and orthogonal vectors. The unitary vectors specify the 2-direction itself, and $\theta$ is the size (analogous to the length of a vector, that is a 1-dimensional direction).

Given two vectors $a$ and $b$ (that we take to be unitary, wlog) we say that the simple bivector $\theta e_1e_2$ is the angle formed by them if
$$ab=e^{\theta e_1 e_2}$$
where we take as a definition
$$e^A=1+A+A^2 /2+\cdots=\lim_N (1+\frac{A}{N})^N$$



Real analysis let us assure that this expression is well defined when $A$ is a 2-blade. In fact, since $e_1e_2e_1e_2=-1$, we can write
$$e^{\theta e_1 e_2}=cos(\theta)+e_1 e_2sin(\theta)$$
where $cos$ and $sin$ are defined by their power series.


Oserve that $b=a e^{\theta e_1 e_2}$ and then
$$b=a(cos(\theta)+e_1 e_2sin(\theta))=a\cdot cos(\theta)+ ae_1e_2 \cdot sin(\theta)$$
and from here we conclude that $a$ is in the plane generated by $e_1$ and $e_2$, since otherwise we would have a trivector in the right hand side of the above equation. Of course, $b$ is also inside the plane.


The fact that $ab=e^{\theta e_1 e_2}$ can be interpreted as follows. We have that
$$b=a e^{\theta e_1 e_2}\cong a(1+\frac{\theta e_1 e_2}{N})^N=a+\frac{\theta}{N}a_{\perp}+\cdots$$
whose meaning is that we arrive to $b$ from $a$ by applying a special infinite process. We need such an infinite process because we are  producing a circle with straight lines operations, as you can observe at this gif:

From this starting point we can prove and define all the usual trigonometry. For example, observe that
$$ab=a\cdot b+a\wedge b=$$
$$=\mu +\lambda e_1e_2$$
and is satisfied that $\mu=cos(\theta)$ and $\lambda=sin(\theta)$. Since $baab=1$ we can show that
$$cos^2(\theta)+sin^2(\theta)=1$$



In this context, we can define $\pi$ as the lowest real positive number $\theta \in \mathbb{R}$ such that
$$e_1 e_2=e^{\frac{\theta}{2} e_1 e_2}$$

Now, natural questions arise: could $\pi$ be the same for every pair of unitary orthogonal vectors $e'_1, e'_2$? Moreover, is the only one?

We can deal with the unicity question by means of real analysis by studying the pair of equations
$$cos(\theta/2)=0$$
$$sin(\theta/2)=1$$
and we will arrive to the usual conclusions.

On the other hand, suppose $e'_1 e'_2$ are other unitary and orthogonal vectors. Then
$$e^{\frac{\pi}{2}e'_1 e'_2}=cos(\pi/2)+e'_1 e'_2 sin(\pi/2)=e'_1 e'_2$$

So $\pi$ is universal for every 2-direction.

From here we can conclude usual facts like
$$e^{\pi i}=-1$$
where we have called $i=e_1e_2$ to resemble the famous expression. To see it, take any unitary $v$ spanned by $e_1, e_2$. And then, since
$$-v=ve_1e_2e_1e_2=ve^{\frac{\pi}{2} e_1 e_2}e^{\frac{\pi}{2} e_1 e_2}=ve^{\pi e_1e_2}$$
and therefore
$$e^{\pi i}=-1$$

Observe that we have used that
$$e^A e^B=e^{A+B}$$
but this is true whenever $AB=BA$.