inscribed angles on circle 
That's basically the problem.  I keep getting $\theta=90-\phi/2$. But I have a feeling its not right. What I did was draw line segments BD and AC. From there you get four triangles. I labeled the intersection of BD and AC as point P. From exterior angles I got my answer. 

One way would be to let $E$ be the center of the circle.  A standard result in geometry tells you that $AEC=2\theta$.  And the two sides $AE$ and $CE$ are of equal lengths, and there are right angles at $A$ and $C$, and the sides $AD$ and $CD$ are also of equal lengths.  So the triangle $EAD$ is right triangle congruent to $ECD$.  One of the angles in that right triangle is $\theta$, so the other is $90^\circ-\theta$.  Therefore what you're looking for is $180^\circ-2\theta$.
