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
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.