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.

 A: I claim that the triangle $\Delta ABC'$ in the diagram below is the desired triangle, and can be constructed geometrically without reference to trigonometry. See below the diagram for a description of the construction.


Construction
The main goal will to be to construct a triangle similar to the one desired, as from there it's easy to scale it up to size.
Start by constructing a $4\times 6$ rectangle $ABCD$, with $A$ in the lower LH corner and $|AB|=6.$ We may then construct an equilateral triangle $CDE$ with the vertex $E$ chosen to be below $CD$ (and in fact below $AB$ as well.) Draw a circle $S$ centered at $E$ that passes through both $C$ and $D$.
Let $M$ be the midpoint of $CD$, and label $A'$,$B'$ as the intersections of the circle $S$ and the lines $AM$, $BM$ respectively. From this we can construct the triangles $A'B'D$, $A'B'C$ which are symmetric across the line $EM$.
We now argue that both of these triangles are, up to reflection, similar to the one desired. Consider the rectangle $R$ formed with $A'B'$ as the lower horizontal segment and with the upper segment on the line $CD$. Since all four vertices lie along lines passing through $M$ and the four vertices of $ABCD$, we conclude that $R$ and $ABCD$ are similar figures. Hence the width and height of the rectangle---and moreover those of the two triangles inscribed within them---are in the ratio $6:4$ as desired.
Now, observe that $A'$ is on the circle $S$ and so $\angle CA'D$ is an inscribed angle of the circle $S$. By the inscribed angle theorem, its angle must be half of $\angle CED$---but $CED$ is an equilateral triangle by construction. Thus $\angle CED=60^\circ$ and so $\angle CA'D=30^\circ$. But this is identical to $\angle B'A'D-\angle B'A'C$. Thus we have constructed two triangles $A'B'C$ and $A'B'D$ which are both similar (up to reflection) to the desired triangle.
From here, all we need to do is rescale one of these triangles to have width $6$. This can be done by drawing a line passing through $A$ and parallel to $A'D$. This intersects $CD$ at a point $C'$. Observe finally that $\Delta ABC'$ and $\Delta A'B'C'$ are similar triangles through the point $M$; since $|AB|=6$, we conclude that $\Delta ABC'$ is the triangle we set out to construct.
A: 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.
A: The only thing I can think of here is to somehow get $\alpha-\beta$ into the picture. One way would be to reflect the triangle about the perpendicular line going through the mid-point of AB as shown here:

So point C' is the reflection of point C. Then angles CAC' and CBC' are both equal to $\alpha-\beta=30$.
I don't know if this helps?
