Construction of a triangle I need to construct a triangle with given information:
$c = 6$, $h = 4$ and $\alpha - \beta = 30º$. 
I'll put approximate result for any clarification.


I have been unable to find a straightforward geometric construction, but an algebraic one is not completely outrageous.

In the diagram, $\tan \alpha  = \frac{4}{x}$ and $\tan \beta  = \frac{4}{{6 - x}}$. From your desired $\alpha  - \beta  = {30^ \circ }$ we get
$$\frac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }} = \tan {30^ \circ }$$
$$\frac{{\frac{4}{x} - \frac{4}{{6 - x}}}}{{1 + \frac{4}{x} \cdot \frac{4}{{6 - x}}}} = \frac{1}{{\sqrt 3 }}$$
$$\frac{{8x - 24}}{{{x^2} - 6x - 16}} = \frac{1}{{\sqrt 3 }}$$
$${x^2} + ( - 6 - 8\sqrt 3 )x + ( - 16 + 24\sqrt 3 ) = 0$$
$$x = \frac{{6 + 8\sqrt 3  \pm \sqrt {{{( - 6 - 8\sqrt 3 )}^2} - 4 \cdot 1 \cdot ( - 16 + 24\sqrt 3 )} }}{{2 \cdot 1}}$$
$$x = 3 + 4\sqrt 3  - \sqrt {73} $$
(The value using $+$ in the $\pm$ is clearly too large.)
That last is not extremely hard to construct, and once you have that the triangle is trivial to construct. Let me know if you need more details on the construction of $x$. ($\sqrt {73}$ can be the hypotenuse of a right triangle with sides $8$ and $3$.)
By the way, the approximate values we get are $x \approx 1.384199$, $\alpha \approx 70.911828^ \circ $, and $\beta \approx 40.911828^ \circ $.
ADDED LATER:
Here is the general solution given $c$ and $h$ for $\alpha  - \beta  = {30^ \circ }$:
$$x = \frac{c}{2} + h\sqrt 3  - \sqrt {{{\left( {\frac{c}{2}} \right)}^2} + {{\left( {2h} \right)}^2}} $$
I left the expression in the square root as it is to make it clear how to construct it as a hypotenuse.
