Let's call the cord ABAB and the centre of the circle CC
Then if you divide the cord in half at MM you get two equal, but mirrored triangles Delta CMA and Delta CMB. These are both rectangular at M. (You should draw this yourself right now !).
angle ACM is half the central angle that was given
(and angleBCMis the other half)
Then sin angle ACM=(AM)/(AC) ->AM=AC*sin angle ACM
Since you know the radius (AC) and the central angle (remember angleACM=half of that), you just plug in these values to get an accurate result for half the chord (so don't forget to double it for your final answer)
