Calculating the radius of the sphere that inscribes a tetrahedron Given the following information about the tetrahedron pictured below, can the radius of the sphere that inscribes the tetrahedron be calculated?
Given:  
Complete information about triangle BCD (side lengths and angles)
The angles formed by BAC, BAD, and CAD

Some physical experimentation with rings representing the circles that inscribe faces BAC, BAD, and CAD (the radii of which can be calculated from the given information) held together on one side at fixed distances (as would be perscibed by the side lengths of face BCD) has shown me that using only the given information I am capable of constructing a unique sphere, though I have yet to find a way to do this mathematically.
 A: Set up the angles at $A$, with angles $\alpha,\beta$ and $\gamma$.  We want to know $AB,AC$ and $AD$ so that $BC=d,CD=b$ and $BD=c$ are the right lengths.  So we want to find $AB=x,AC=y$ and $AD=z$ for which 
$$x^2-2xy\cos\alpha+y^2=d^2\\
y^2-2yz\cos\beta+z^2=b^2\\
z^2-2xz\cos\gamma+x^2=c^2$$
I think three degree-2 equations will, in general, have eight ($=2^3$) solutions.  
I entered it into Maple, which gave me a degree-8 polynomial for $x$.  Here is how you might do it:  


*

*Subtract equation (2) from equation (1).  The result is linear in $y$.  

*Multiply equation (1) by $(2x\cos\alpha-2z\cos\beta)^2$.  That lets you substitute $y$ altogether, leaving $P(x,z)=0$, a degree-4 polynomial in $x$ and $z$.  

*Regard equation (3) as a quadratic in $z$.  It has two solutions $z_1(x)$ and $z_2(x)$.  The only things I need from this is $z_1(x)+z_2(x)=2x\cos\gamma$ and $z_1(x)z_2(x)=c^2-x^2$.  

*$P(x,z_1)P(x,z_2)=0$ is a degree-8 polynomial in $x,z_1$ and $z_2$, that is symmetric in $z_1$ and $z_2$.  So it can be written in terms of $x,z_1+z_2$ and $z_1z_2$.  Now it can be written as a degree-8 polynomial in just $x$.  

*The original equations have a symmetry, that $(x,y,z)$ can be replaced by $(-x,-y,-z)$.  Therefore, the degree-8 polynomial is even, and is a degree-4 polynomial in $x^2$.


EDIT 2:
Here is a formula if all six edges are known, and the three angles mentioned above.
Let R be the circumcentre.  It is the intersection of the planes that are the perpendicular bisectors of the edges.  Any three of these planes will do so long as all four vertices are involved.
Put $A$ at the origin.  The equation of the plane bisecting $AB$ is $$2\vec{R}.\vec x=\vec x.\vec x$$
Let $M$ be the matrix whose rows are $\vec x,\vec y,\vec z$, and $V$ the column vector whose entries are $x^2,y^2,z^2$.  Then $2M\vec R=V$, and $\vec R=\frac12M^{-1}V$.  Dot product $\vec R$ with itself to get $R^2$, the square of the circumradius
$$R^2 = \frac14V^T(M^{-1})^TM^{-1}V= \frac14V^T(MM^T)^{-1}V\\
=\frac14
(x^2,y^2,z^2)
\left(\begin{array}{ccc}
x^2&\vec x.\vec y&\vec x.\vec z\\
\vec y.\vec x&y^2&\vec y.\vec z\\
\vec z.\vec x&\vec z.\vec y&z^2\end{array}\right)^{-1}\left(\begin{array}{c}x^2\\y^2\\z^2\end{array}\right)$$
Recall that $\vec x.\vec y=xy\cos\alpha,\vec y.\vec z=\cos\beta,\vec z.\vec x=\cos\gamma$, and I think it works out to
$$\frac{x^2\sin^2\beta+y^2\sin^2\gamma+z^2\sin^2\alpha+2xy(\cos\beta\cos\gamma-\cos\alpha)+2xz(\cos\alpha\cos\beta-\cos\gamma)+2yz(\cos\alpha\cos\gamma-\cos\beta)}
{4(1+2\cos\alpha\cos\beta\cos\gamma-\cos^2\alpha-\cos^2\beta-\cos^2\gamma)}$$
That can be written just in terms of $b,c,d,x,y,z$ by using my first three equations.  I don't know whether it can be written just in terms of $b,c,d,\alpha,\beta,\gamma$, which was your original question.
EDIT 3:
The denominator is 
$$16\sin(\frac{\alpha+\beta+\gamma}2)\sin(\frac{\alpha+\beta-\gamma}2)\sin(\frac{\alpha-\beta+\gamma}2)\sin(\frac{-\alpha+\beta+\gamma}2)$$
while the numerator is 
$$\frac{(xb+yc+zd)(xb-yc+zd)(xb+yc-zd)(-xb+yc+zd)}{b^2c^2c^2}$$
A: The inscribed circles on the faces of the tetrahedron don’t have a particularly simple relationship to its inscribed sphere. A way to view one of these circles is as the intersection of an elliptical cone with the face. This cone is tangent to the other three faces and to the inscribed sphere. I don’t believe that the axes of these cones pass through the sphere or circle centers, so it doesn’t seem to me that examining them really gets you any closer to finding the inscribed sphere.  
Based on your comments it seems that you’re really interested in finding $A$ by somehow using the inspheres of a bunch of tetrahedrons that share this vertex. All of the ways that I can imagine to compute the insphere either require knowing $A$ or computing things like the face normals that are tantamount to knowing $A$. I don’t really see how working with inspheres would lead to anything more tractable than a direct computation.  
Unfortunately, the information you have about the tetrahedron doesn’t lend itself to a straightforward solution: You have some of the side lengths and angles, but not the ones that could be used in the usual side-angle-angle formulas. Michael gives a direct approach to this in his answer, essentially turning the problem into one of finding the intersection of three quadrics. This is likely to have multiple solutions, but it seems to me that you should be able to disambiguate by considering the other tetrahedra in your larger problem.  
Bringing either projective geometry or spherical trigonometry to bear on the problem might be useful. For example, given the three known angles at $A$, one can easily construct three rays from it with those angles between them. The problem of placing $A$ is then equivalent to finding a plane such that the pairwise distances between its intersections with those rays match the known edge lengths. Using spherical trigonometry, the dihedral angles of the faces meeting at $A$ can be computed, and it should be possible to develop constraints on the unknown angles and relationships among them. (There is, for instance, a tetrahedral counterpart of the Law of Sines that can relate adjacent angles at a vertex.)
