Given that a triangle has edges of ratio 2 : 3 : 4, the task is to determine the three angles, say in degrees. I started by drawing 4 cm segment on the paper, then drew perpendicular segments of lengths 2 and 3 cm going up from the endpoints of the original segment. With a compass I drew arcs from the endpoints of the perpendicular segments, and their intersection allowed me to determine a triangle s.t. its edges meet the 2 : 3 : 4 (cm) ratio requirement. Visually:
As happy as I am for being able to construct the triangle, I don't quite know how to proceed. I feel that the 2-4 angle (sorry I forgot to label) is 45°, the 3-4 one 30°, leaving 105° for the 2-3 angle. But I need some guidance as to how to approach actually proving those (or if those guesses are wrong, how to determine the actual angles). Thanks in advance for any hints/suggestions.
Let your triangle be $ABC$, with $A$ at the bottom left, $B$ at the bottom right, and $C$ on top.
Drop a perpendicular from $C$ to $AB$, meeting $AB$ at $P$.
Let $AP=x$. Then $BP=4-x$. Let $CP=h$.
By the Pythagorean Theorem on the two smaller triangles, we have $$x^2+h^2=4 \quad\text{and}\quad (4-x)^2+h^2=9.$$ Expand and subtract. We get $16-8x=5$ and therefore $x=\frac{11}{8}$.
It follows that the cosine of the angle at $A$ is equal to $\frac{11/8}{2}$, that is, $\frac{11}{16}$. Thus (calculator) the angle at $A$ is about $46.57^\circ$.