Assuming the unlabeled line is parallel to the x-axis you could use the law of cosines.
Let us label each point $A(x_1,y_1)$, $B(x_2,y_2)$ and $C(x_1+r,y_1)$ where $r$ is the radius and $x_1,x_2,y_1,y_2$ are positioned as in you diagram.
The distance between $A$ and $B$ is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ and the distance between $B$ and $C$ is $\sqrt{(x_2-x_1-r)^2+(y_2-y_1)^2}$.
Now, using the law of cosines which is $a^2=b^2+c^2-2bc\cos A$ we get $(x_2-x_1-r)^2+(y_2-y_1)^2=r^2+(x_2-x_1)^2+(y_2-y_1)^2-2\cdot r\cdot \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\cos A$
By solving for $\cos A$ we get: $\cos A={(x_2-x_1-r)^2+(y_2-y_1)^2-r^2-(x_2-x_1)^2-(y_2-y_1)^2\over -2\cdot r\cdot \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$
Which simplifies to: $\cos A={x_2-x_1\over \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$
Therefore the angle is: $A=\cos ^{-1} \left( {x_2-x_1\over \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}\right )$
I'm sorry if this was overly complicated for the question. There probably is some simpler way to do it, but it will look a lot less messy once you plug in the numbers.
Note: Please feel free to comment if I have made any careless mistakes.