What is the angle of a line that is tangent to a circle and passes through an arbitrary point? I made this illustration to clarify the question:

How would I find the angle of this line relative to the bottom line? If the boxes have length x instead of 1, what is the generic solution?
 A: If the boxes have length $x$, then we have a similar figure, so that the angle must remain the same.  It suffices, then, to find the angle for $1$.  
Here's a somewhat "brute force" approach, but it gets the job done:


*

*Draw a radius from the center (of the circle) to the point of tangency.  This radius intersects the line at a right angle.

*Draw a line from the lower-left corner to the center 

*Draw a line from the center horizontally to the left, to the point where the left side is tangent to the circle.
We now have two right triangles that fall "above" the original diagonal line.  The angle we want is the complement of the sum of the two angles.
Lower triangle: the hypotenuse has length $\sqrt{(1/2)^2 + (3/2)^2} = \sqrt{10}/2$.  The opposite leg has length $1/2$.  The angle of the desired corner is thus
$$
\sin^{-1}\left(\frac{1/2}{\sqrt{10}/2}\right) = 
\sin^{-1}(1/\sqrt{10})
$$
Upper triangle: the other triangle is congruent.  The other angle is the same.
The desired angle is thus
$$
90^\circ - 2\sin^{-1}(1/\sqrt{10}) \approx 53.1^\circ
$$
A: 
AB = $x + \frac{x}{2} = \frac{3x}{2}$
$\angle AOB = tan^{-1}\frac{x}{2}/\frac{3x}{2} = \tan^{-1}(\frac{1}{3})$ and
$\angle CAD = \pi/2 - 2 \angle AOB$
A: we will use the fact that the two tangents drawn from a point are equal and line connecting the center and the point bisects the angle between the tangents. the left  side of the rectangle is one of the tangents and is of length $\frac32$ the half angle between the tangents is $\tan^{-1}(1/3) = 18.43^\circ.$  the angle the right tangent makes with the width of the rectangle is $90^\circ - 2*18.43 = 53.13^\circ.$
A: 
When 1 is x,
$\theta$ is given by $\tan \theta = \frac {0.5x}{0.5x + x} = \frac {0.5}{1.5}$.
