TITLE: Construction of a triangle
QUESTION [8 upvotes]: 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.

REPLY [4 votes]: 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.