Shenzhen bc recruitment

∵BP is the tangent of⊙ O,

∴OB⊥BD，

That is ∠ OBD = 90,

∫∠A = 30，

∴∠BOC=2∠A=60，

OB = OC，

∴△OBC is an equilateral triangle,

∴OB=OC=BC=2，

∴BD=OB？ tan60 =23，

∴S shadow =SRt△OBD-S sector OBC =12× 2× 23-60360× π× 22 = 23-23 π.

So the answer is: 23-23 π.