Axiomatize inner product spaces in terms of angles? Inner product space is a vector space $V$ over a field $F$ together with an inner product $P:V\times V\to F$ that satisfies the inner product axioms. Let $V_{-0} = V \setminus \{0\}$.
This inner product induces an angle $\angle (v,w) = \arccos \left(\frac {P(v,w)}{||v||\cdot ||w||}\right)$, where $\angle : (V_{-0})\times (V_{-0}) \to [0,\pi]$.
I am wondering if there is an axiomatization of an "angled vector space" $(V,\angle)$ with $\angle:(V_{-0})\times (V_{-0})\to [0,\pi]$, such that 

"angled vector space axioms". What are a set of axioms on a pair $(V,\angle)$ such that they are satisfied if and only if there exists an inner product space that induces $\angle$?

 A: There may be more "natural" axiomatizations, but you can get an axiomatization that works rather straightforwardly by just reconstructing an inner product (up to scaling) from its angle function (or even just from knowing which vectors are orthogonal).
Indeed, suppose $\angle$ is the angle function of an inner product.  We say $v,w\in V$ are orthogonal if either $\angle(v,w)=\pi/2$ or either $v$ or $w$ is $0$.  Then note that $\|v\|=\|w\|$ iff $v+w$ and $v-w$ are orthogonal.  So, we can recover when two vectors have the same norm, which means we can recover the norm itself up to scaling.  We can then recover the inner product by polarization.
So here, then, are some axioms you can use for $\angle$ which are equivalent to it coming from an inner product.  For $v,w\in V$, say that $v\sim w$ if $v+w$ and $v-w$ are orthogonal.


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*Axiom 1: $\sim$ is an equivalence relation, and for any nonzero $u\in V$ and any $v\in W$, there is a unique $c\geq 0$ such that $cu\sim v$.


Assuming Axiom 1, for any nonzero vector $u$, we write $\|v\|_u$ for the unique $c\geq 0$ such that $cu\sim v$.  We also write $P_u(v,w)=\frac{1}{4}(\|v+w\|_u^2-\|v-w\|_u^2).$


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*Axiom 2: For any nonzero $u\in V$, $P_u$ is an inner product on $V$, and $\angle (v,w) = \arccos \left(\frac {P_u(v,w)}{P_u(v,v)^{1/2} P_u(w,w)^{1/2}}\right)$ for all nonzero $v,w\in V$.


Obviously these two axioms imply that $\angle$ comes from an inner product (namely $P_u$ for any nonzero $u$; I leave the case $V=0$ to the reader).  Conversely, if $\angle$ comes from an inner product $Q$, then Axiom 1 is satisfied since $v\sim w$ means $\|v\|_Q=\|w\|_Q$, and Axiom 2 is satisfied since $P_u$ is just $Q$ rescaled so that $u$ is a unit vector.
A: I think that the following should work, but I did not have the time to check all the details.
We use the following, naturally occuring axioms:


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*Nontriviality: $\angle(v,w) = 0$ iff $v = \lambda \, w$ for some $\lambda > 0$

*Summing of angles: $\angle(v,w) + \angle(-v,w) = \pi$

*Symmetry: $\angle(v,w) = \angle(w,v)$

*Triangle-Equality (triangle with vertices $0$, $v$, $w$): $\angle(v,w) + \angle(-v,w-v) + \angle(-w,v-w) = \pi$

*Scaling: $\angle(v,w) = \angle(\alpha \,v, \beta\,w)$ for all $\alpha, \beta > 0$

*Iterated law of sines:
$$\frac{\sin(\angle(-v,w-v))}{\sin(\angle(-w,v-w))} = \frac{\sin(\angle(-u,w-u))}{\sin(\angle(-w,u-w))} \cdot \frac{\sin(\angle(-v,u-v))}{\sin(\angle(-u,v-u))}$$
whenever all denominators are nonzero.


Now, we define a norm on $V$ by the following procedure: fix $v \in V_0$ and define $\|\lambda \, v\| = \lambda$ for all $\lambda \in \mathbb R$. 
For an arbitrary vector $w$ which is not in the span of $v$, consider the triangle with vertices $0$, $v$, $w$. From this triangle, we know the length of one side and all three angles. Hence, the triangle can be constructed in the Euclidean plane and we can measure the length of $w$. This leads to
$$
\|w\| = \frac{\sin(\angle(-v,w-v))}{\sin(\angle(-w,v-w))} \ne 0.$$
Now, we have to check the axioms of a norm.


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*$\|w\| = 0$ implies $w = 0$ per construction.

*$\|\alpha \, w\| = \alpha \, \|w\|$ follows from a two-dimensional consideration.

*Triangle inequality follows from triangle inequality in $\mathbb R^2$.


Finally, the parallelogram identity should follow from a three-dimensional consideration.
