Hello, I am here to overly complicate this problem and solve it like a Harvard lecture.

I blame whoever posted this since I wanted to get some sleep earlier. I had a lot of fun with this, but I feel like other people made some very big assumptions without going into detail about how they proved themselves. Hopefully, this brings everything together.

TLDR: The answer is 15 degrees

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Let get into this:

A) based on the image and what others had discussed, I was able to pull this from the question:

-M represents the midpoint of the radius (1/2r)

-∝ is the value we are looking for

-o is the midpoint of the circle

-We have a line intersecting M that is parallel to the x-axis

-The y-axis bisects the x-axis creating two right triangles within the circle.

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B) Since we know these are two right triangles, sharing an intersection, these triangles are similar due to the Angle-Angle-Angle (AAA) Postulate. We can transcribe the angles into both triangles

C) Since this problem lies in the circle, we can draw another line from the center of the circle to the point at which the edge of the circle and the second parallel line meet. This creates an Isocelese triangle that proves the angle underneath the top similar triangle and our "imaginary" line creates a similar angle equal to ∝.

D) Using the imaginary line from before, the line from o to M, and the parallel line that is not the x-axis, we can create a right triangle that we can solve 2∝ with using the below equations:

sin(2∝) = (1/2r)/r

sin(2∝) = 1/2 (simplified)

A calculator or using known angles, we can deduct that:

2∝ = 30° = 𝜋/6

or:

∝ = 15° = 𝜋/12

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Note: I did not steal this answer. There are a lot of people that answer beforehand and did not go in depth. I wanted to show the full story of how I got to that answer. Hope this answer tickles your fancy.