Does anyone know how to find the diameter? It appears the lack of sleep is getting to me.

Chord BD=8, and minor arc BA=60 degrees

If I remember geometry, BD is perpendicular to BA, so the diameter (DA) is 8/(cos(30))

To confirm his answer in a complicated round about way:

Line Seg BC would create an iscoceles triangle because two sides are equal to the radius. Using laws of sine

8/sin (120) = r/sin30

therefore

d=2(8sin(30)/sin(120))

If you evaluate you get the same number approximately 9.237 or

16/(sqrt(3) or rationalized 16sqrt(3)/3

Hope I helped haha

Thanks, I read what you said and had one of those slow… ohhh yeah moments.

Another much simpler way is to form a 30-60-90 triangle BDA which allows you to find DA by dividing the segment BD by 2, and then times that by root 3 to find the length BA, which is the diameter.