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.)
 A: 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 $90^\circ$.
A: Suppose your third plane cuts plane 1 along line $OR$ and plane 2 along line $OQ$. Let $\alpha$ and $\beta$ be the two angles which $OR$ and $OQ$ form with $AB$ and choose point $R$ such that $OR=1$. Let then $P$ be the projection of $R$ onto $AB$ and choose point $Q$ so that $PQ\perp AB$.
From elementary trigonometry applied to triangle $OPR$ and $OPQ$ we get $PR=\sin\alpha$ and $PQ=\cos\alpha\tan\beta$. We can then compute $QR^2$ in two ways, applying the cosine rule to triangles $OQR$ and $PQR$ (notice that $OQ=\cos\alpha/\cos\beta$):
$$
QR^2
=1+{\cos^2\alpha\over\cos^2\beta}-2{\cos\alpha\over\cos\beta}\cos\phi
=\sin^2\alpha+{\cos^2\alpha\over\cos^2\beta}\sin^2\beta-2{\cos\alpha\over\cos\beta}\sin\alpha\sin\beta\cos\theta.
$$ 
From that, you can express $\cos\phi$ in terms of $\alpha$, $\beta$ and $\theta$:
$$
\cos\phi=\cos\alpha\cos\beta+\sin\alpha\sin\beta\cos\theta.
$$

