Construction with compass and straight edge of a triangle. How can I construct a triangle given the altitudes of side $a$ and side $b$ and the angle $\beta$ at vertex $B$?
 A: Let $p$ be the given altitude to the side $a$, and let $q$ be the given altitude to the side $b$. 
We produce an algebraically motivated solution.  The idea is that we first produce a triangle similar to the desired triangle, and then scale it to the desired size.
Draw an arbitrary line segment, which we call $A'B'$. At $B'$, draw a line $\ell$ that makes angle $\beta$ with line $A'B'$. Drop a perpendicular from $A'$ to the line $\ell$, meeting $\ell$ at say $M'$ 
Let $p'$ be the length of the just drawn perpendicular. Construct a line segment of length $q'=\frac{p'}{p}q$.  This can be done by compass and straightedge. 
Now draw the circle $X$ with centre the midpoint of $A'B'$, and diameter $A'B'$. Draw the circle with centre $B'$ and radius $q'$. These circles meet at a point $N'$ such that the altitude from $B'$ to $A'N'$ has length $q'$. Draw the line $A'N'$, and let it meet $\ell=B'M'$ at $C'$.
The triangle $A'B'C'$ has the shape that we want, except for being wrong by a scaling factor of $p'/p$. Scale it up, by multiplying the sides by $p/p'$. This can be done by straightedge and compass.   
A: Let the lengths of the altitudes to the sides $a$ and $b$ be $p$ and $q$.

*

*Draw a line $\ell_C$. Pick a point $B$ on it, and draw line $\ell_A$ through $B$ at angle $\beta$ to $\ell_C$.
Lines $\ell_A$ and $\ell_C$ form two of the sides of the triangle.


*Draw a line parallel to $\ell_A$ at distance $p$, meeting $\ell_C$ at point $A$.
Now the altitude from $A$ to $\ell_A$ has length $p$.


*Draw a circle of radius $q$ centered at $B$. Draw line $\ell_B$ through $A$ tangent to this circle, meeting $\ell_A$ at point $C$.
Now the altitude from $B$ to $\ell_B$ has length $q$.
Thus, $\triangle ABC$ is the desired triangle. (This is essentially the same as André Nicolas's construction with the modification I suggested in a comment.)
