Elementary 3D geometry This is surely trivial, but my old brain can't remember how to do it.
Assume a plane.
A second plane intersects, forming line $AB$. The angle of intersection is $\theta$.
A third plane intersects, crossing $AB$.
At the point of intersection, we can project a triangle in Plane 3 whose sides are established by the intersections of Plane 3 with Plane 1 and Plane 2 respectively.
If Plane 3 intersects $AB$ at 90 degrees, the vertex of the triangle (at $AB$) has angle $\theta$ -- this should be obvious, and if not, I've described the problem wrong.
So suppose the intersection of Plane 3 is not 90 degrees but something else, $\phi$.
What is the angle of the vertex of the triangle?
(Suggestions for stating this more clearly are very welcome.)

I don't think your problem is properly defined. The vertex angle of the triangle is not completely determined by the angle $\phi$. Saying it another way ... there are infinitely many planes that make an angle $\phi$ with the line $AB$, and different planes will give you different triangle vertex angles.
Here's an example ...
Suppose plane $P1$ is the plane $y=0$ and plane $P2$ is the plane $x=0$. Then the intersection line $AB$ is the $z$-axis, and $\theta = 90^\circ$. Take $\phi = \tan^{-1}\tfrac34$.
One possible plane $P3$ that makes an angle $\phi$ with $AB$ is the plane $4x-3z=0$. It gives a triangle vertex angle of $90^\circ$. 
Build another plane $P4$ by rotating $P3$ around $AB$ by $45^\circ$. Clearly $P4$ makes the same angle $\phi$ with $AB$, but I think it's obvious geometrically that it gives a triangle vertex angle of less than $90^\circ$.
If you don't believe the "obvious" claim, you can confirm by computations. The plane $P4$ has equation $8x+8y-9z=0$. Its intersection with $P1$ is the line $8x-9z=0$, and its intersection with the $P2$ is the line $8y-9z=0$. You can calculate the angle between these two lines and confirm that it's less than $90^\circ$.
