I interpreted your query in two methods. If by $frac{π}{2}$ you imply the arc distance traveled, then right here is one reply.
The three segments space all tangent to the circle. Therefore, that factors at which they meet are the highest vertex of isosceles triangles. Check out the correct triangle fashioned by the radius touching the underside line and the gap left (or the one traveled) by the circle. We take $arctan(angle)=frac{radius}{distance}$, isolate the angle and double it, as a result of it offers us one half of the isosceles triangle. If we try this for each backside angles, this provides us a formulation for the highest fashioned by the tangent traces. $a=π-arctan(frac{1}{2π-d})-arctan(frac{1}{d})$ the place d is the arc distance traveled. If we graph such operate, it turns into apparent that the utmost of the curve within the most angle potential, which is midway $frac{π}{2}$, that’s about 1,91rad. The curve, which supplies us the angle linked to the gap traveled “d” factors downwards, so on either side of its most, the angle is smaller.
If by $frac{π}{2}$ you imply the worth of the angle itself, then it offers us about 0,69rad, evaluating a proper angle, of worth 1,57, then it’s smaller than 90 levels.
Edit: Sorry, I made a mistake guys. The equation describing the angle on the prime can be $a=π-2cdot arctan(frac{1}{2π-d})-2cdot arctan(frac{1}{d})$ giving us 1,59rad at $frac{π}{2}$, which itself is 1,57rad. My remaining reply is that angle $alpha$ is obtus. Actually do not know the place I gathered that 0.69rad.
PS: Thanks for the welcome too! I am going to do higher subsequent time.
