Question: > 
> $\mathbf{a}\times \mathbf{b}$ follows the right hand rule? Why not left hand rule? Why is it $a b \sin (x)$ times the perpendicular vector? Why is $\sin (x)$ used with the vectors but $\cos(x)$ is a scalar product?
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So why is cross product defined in the way that it is?
I am mainly interested in the right hand rule defintion too as it is out of reach?

Answer: 
The cross product originally came from the *quaternions*, which extend the complex numbers with two other 'imaginary units' $j$ and $k$, that have noncommutative multiplication (i.e. you can have $uv \neq vu$), but satisfy the relations

$$ i^2 = j^2 = k^2 = ijk = -1 $$

AFAIK, this is the exact form that Hamilton originally conceived them. Presumably the choice that $ijk = -1$ is simply due to the convenience in writing this formula compactly, although it could have just as easily been an artifact of how he arrived at them.

Vector algebra comes from separating the quaternions into scalars (the real multiples of $1$) and vectors (the real linear combinations of $i$, $j$, and $k$). The cross product is literally just the vector component of the ordinary product of two vector quaternions. (the scalar component is the negative of the dot product)

The association of $i$, $j$, and $k$ to the unit vectors along the $x$, $y$, and $z$ axes is just lexicographic convenience; you're just associating them in alphabetic order.
This may be a bit too deep but let $V$ be a finite dimensional vector space with basis $v\_1,...,v\_n$. We say $(v\_1,...,v\_n)$ is an oriented basis for $V$. We can define an equivalence class on orientations of $V$ by $[v\_1,...,v\_n] \sim [b\_1,...,b\_n] \iff [v\_1,...,v\_n] = A[b\_1,...,b\_n]$ (where $A$ is the transition matrix) and $\textbf{det}(A)>0$. Therefore, orientations are broken up into two classes, positive and negative. If we let $\textbf{e}^1,...,\textbf{e}^n$ be the standard basis for $\mathbb{R}^n$ then if we wish to check that $[b\_1,...,b\_n] \sim [\textbf{e}^1,...,\textbf{e}^n]$ then we simply have to look at;

$$\textbf{det}\left(\begin{bmatrix} b\_1 & b\_2 & \cdots & b\_n \end{bmatrix}\right) >0$$

Since $A = [b\_i]$ is the change of basis matrix. 

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Edit (1): Above gives a generalization of determining orientation. In your case you would have $E=[\textbf{e}^1,\textbf{e}^2, \textbf{e}^3]$ which gives positive orientation and the ''right hand rule'' should be thought of as a geometric property that one can check without knowing all the information I presented above. 

Edit (2): In regards to your question about why $\sin \theta, \cos \theta$ are used in certain situations is also because of geometry. The area of a parallelogram is base $\times$ height $= |\vec{a} \times \vec{b}| \sin \theta$ where $\vec{a}, \vec{b}$ span the parallelogram.