+1768 Two points on a circle of radius $1$ are chosen at random. Find the probability that the distance between the two points is at most $1.5.$
+129850 
Since all rotations are the same, let A be any point on the circle
AC = 1.5 = AB
DA = 1 = DB
Using the Law of Cosines
BA^2 = DA^2 + DB^2 - 2 (DA * DB)cos (ADB)
1.5^2 = 1^2 - 1^2 - 2 ( 1 * 1) cos (ADB)
[ 2.25 - 1 -1 ]/ [ -2 ] = cos (ADB)
.25 / -2 = cos (ADB)
-(1/8) = cos (ADB)
arccos (-1/8) ≈ 97.18° = ADB = ADC
Probability ≈ [ 2 measure ADB ] / 360 = [2 * 97.18 ] / 360 ≈ 54%
