How does the angle between tangent vectors of a curve at different points and a given fixed uniform vector field change on rotation of curve?

Question

I have to evaluate an integral $\int B\ dl \sin\theta$ for quarter $CD$ of the given ring of radius $r$.
[$\theta$ being angle between tangent vector($d\vec l$) and $\vec B$]

1. In the case where $\vec B$ was parallel to plane of ring:

(the arrows on the ring represent direction of tangent vectors to be taken)
The integral becomes $\int B\ dl \sin\theta=\int B\ r\sin\theta \ d\theta$ as change in $\theta$ is equal to in angle subtended by $dl$
The limits of integration for $\theta$ from C to D were $0$ to $\frac{\pi}{2}$.
This I figured easily using geometry.

2. In case when $\vec B$ was perpendicular to plane of ring:
   Value of integral becomes $B\sin(\pi/2)\int dl$ since $\vec B$ is perpendicular to tangent everywhere.

3. In case when ring is rotated about its diameter by an angle such that
   Ring is now inclined at an angle $\phi$ to vector field $\vec B$
   This was the general case but I could not imagine nor draw a diagram to find the values of angle between $d\vec l$ and $\vec B$.
   
   For the third case

• Can you provide a diagram or explanation of some sort to show how $\theta$ changes from $C\rightarrow D$?
• Is the angle between area($\vec A$) and $\vec B$ equal to $\cfrac{\pi}{2}-\phi$?
• What are the limits of integration and how will I relate $r\ $,$dl\ $,$B\ $,$\phi\ $,$\theta\ $ to compute the integral ?