Expert Help with Non-Right Triangle Equations

One of the largest issues in ancient mathematics was accuracy-nobody had calculators that went out ten decimal places, and accuracy generally got worse as the numbers got larger. The famous Eratosthenes experiment. Given two similar triangles, one with small measurements that can be accurately determined, and the other with large measurements, but at least one is known with accuracy, can the other two measurements be deduced? Explain and give an example. The similarity of triangles gives rise to trigonometry. 
How could we understand that the right triangles of trigonometry with a hypotenuse of measure 1 represent all possible right triangles? Ultimately, the similarity of triangles is the basis for proportions between sides of two triangles, and these proportions allow for the calculations of which we are speaking here. The similarity of triangles is the foundation of trigonometry.

What is the inradius of the octahedron with sidelength a?
1) Determine the height of one of two constituent square pyramids by considering a right triangle, using the fact that the height of the equilateral triangle is ${\displaystyle a\sqrt{\frac{3}{2}}}$. (Incidentally, by this you obtain the circumradius of the Octahedron.)
2) Now cut the (solid) octahedron along 4 of these latter heights, i.e. across two non-adjacent vertices and two midpoints of parallel sides of the ''square base''. (Figure 1)
Consider the resultant rhombic polygon, whose sidelength is ${\displaystyle a\sqrt{\frac{3}{2}}}$. Choose a convenient pair of adjacent corners A and B, s.t. the angle in A is obtuse. Then the insphere touches AB, say in M.
So the question becomes: Is there a still simpler proof?

Given ${\displaystyle b=2,a=3}$, and ${\displaystyle B=40}$ (degrees), determine whether this information results in one triangle, two triangles, or no triangle at all. Any resulting triangles should be resolved.