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.

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 
$$\frac{(xb+yc+zd)(xb-yc+zd)(xb+yc-zd)(-xb+yc+zd)}{b^2c^2c^2}$$
