23.在三棱锥O−ABCO-ABCO−ABC中,∠AOB=∠BOC=∠AOC=60∘\angle AOB=\angle BOC=\angle AOC=60\circ∠AOB=∠BOC=∠AOC=60∘,则直线OAOAOA与平面BOCBOCBOC所成角的正弦值为 ___63\frac{\sqrt{6}}{3}36___.