Why is cross product defined in the way that it is? 
$\mathbf{a}\times \mathbf{b}$ 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?

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?

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. 
