How do we know there is a concept of angle even beyond $3D$ space? I was watching a video on Linear Algebra, the trainer describes how the angle ($\theta$) between two vectors $a^{T} = [a_1, a_2]$ & $b^{T}= [b_1, b_2]$ in $2D$ space is $\theta = cos^{-1}\{\frac{dotProduct(ab)}{|a||b|} \}$ where denominators in the right hand side are magnitudes. The trainer generalizes this equation to $3D$ which was digestible to me as I studied & derived it back in school days but what startled me is the trainer generalized it to $N$-dimensional space as well. Now at least I can't comprehend any space beyond $3D$ then how the concept of angle lurking there? Or Is there something called 'angle' beyond $3D$ space? If it is there how do we know? 
how do we know there is concept of angle even beyond $3D$ space?
 A: In higher dimensions, you simply take that formula as the definition of the angle between two vectors.
You can do this since the Cauchy–Schwarz inequality says that $|a\cdot b|\le |a||b|$, so that you only ever get $\arccos x$ for $x$ inside the allowed interval $[-1,1]$.
A: We define an angle as:
Let $\vec u, \vec v \in \mathbb R^{m×1}$, then $$\theta= \arccos(\frac {\vec u \cdot \vec v}{\|\vec u\| \| \vec v \|})$$
Obviously, as you can see $\langle u,v \rangle = \vec u^t \vec v$ is defined since both $\vec u, \vec v \in \mathbb R^{m×1}$. Similarly, you can verify that $$\|\vec x\|, \forall \vec x \in \mathbb R^{m×1}$$ is defined. Furthermore, notice: $\|\vec x\|, \forall \vec x \in \mathbb R^{m×1}=\alpha$, where $\alpha \in \mathbb R $
Note: $\vec u$ and $\vec v$ must not be $\vec 0$
Therefore, we can "meaningfully" speak about the angle between any two $m$-dimensional vectors. 
What does an angle mean, when it is extended beyond the routine 3-D space?
It means exactly what it means in 3-Dimensions, with the only difference that we can't visualize them. 
In essence, we have angles beyond 3-D because that is exactly how we define an angle (refer to the beginning of the answer), especially in contemporary mathematics. Obviously, then, for an ancient greek mathematician working with a straight edge and a compass, an angle would be something that only exists 3-D, but we have come far from that sort of primitive mathematics. 
A: $\DeclareMathOperator{\span}{span}$Other answers have already mentioned that the inner product formula for the angle is usually taken as the definition. But to get at why we can do this and why angles should still work as expected, we have to use a different approach.
Suppose we're working in $\mathbb R^n$, $n\geq3$, and we want to look at angles. Suppose we have two independent vectors $a,b$ and visualise them as arrows coming from the origin. It should be fairly obvious that $\span(a,b)$ precisely defines a plane, and both $a,b$ lie on it. Since this subspace of $\mathbb R^n$ is spanned by two linearly independent vectors (that is to say, they form a basis), the dimension of this subspace is $2$. Hence the concept of the angle between $a,b$ can be defined as the angle between them when only looking at the two dimensional subspace $\span(a,b)$, which "works" just like $\mathbb R^2$. In formal terms, there's an orthogonal transformation taking $\span(a,b)$ to $\mathbb R^2$, and we just say the angle between $a,b$ inside $\mathbb R^n$ is the angle between them inside $\span (a,b)$, which we take to be the angle between them in $\mathbb R^2$. This gives a natural way of generalising angles to higher dimensions by looking at subspaces.
