(2013?

Solution: ①In the process of B→A, when OD=3+33-3=33, ⊙D and ⊙O are circumscribed, at this time BD=OB+OD=3+33, < /p>

∵∠C=90゜, ∠A=30°, BC=6,

∴AB=2BC=12, AC=BCtan30°=63,

∴∠B=60°,

∴BH=BC?cos∠B=6×12=3, CH=BC?sin∠B=33,

∵OB= 3.

∴O coincides with H,

That is, OC⊥AB, OC=33,

∴∠BOC=90°-∠B=60° ,

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