How can I find the radius of curvature of a pipe when given the angle?

Question

I am given a pipe with a 3mm diameter with walls of nearly infinite thinness (so the impact is not affected the the thickness of the pipe) that has something travelling down the center of it in a line. How can I find the minimum radius of curvature of the center line of this pipe such that when the line intersects the wall it will not have an angle greater than 69 degrees?

I have tried making a triangle that connects the point where the curve starts, the impact point, and the center of the curve, but that only tells me the angles of that triangle, leaving the sides all unknown.

I have found that the length of the outer arc within the triangle has a length of 1.2 times the radius, but again, that isn't helpful. I have no other ideas about how I would even proceed with this and searching online for something I can extrapolate the answer from has been fruitless.

The goal is to then find the radius of the circle that is created by curving at that rate.

Added explanation from comments: The problem can essentially be boiled down to: Two lines are travelling parallel to each other 1.5 mm apart along the x-plane. At some arbitrary point, the line below curves up and intersects the line above it, creating an angle of 69 degrees from the tangent of the curving line. If the line continues curving at the same speed to create a circle, what is the radius of that circle (i.e. the radius of curvature).

Answer 1

Here's some visual guidance:

We thus must have that

$$ R = 1.5 + R \cos{69º} $$

Answer 2

The circular arc before hitting the wall describes a sector of a circle with radius $r$ and angle $\theta$. One radius of the sector is straight vertical, and the other intersects the wall perpendicular to the wall. We therefore have $$\theta + 90 + (90-69) = 180$$ and so $\theta = 69$ degrees. Finally $$r\cos\theta = r-1.5$$ or $$r = \frac{1.5}{1-\cos 69^\circ} \approx ~2.34\ \mathrm{mm}$$

(Sanity check: if $69^\circ$ were $90^\circ$ instead, the above would give $r=1.5\ \mathrm{mm}$, which is as expected.)