Let's call the cord #AB# and the centre of the circle #C#
Then if you divide the cord in half at #M# 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 #angleBCM#is 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)
