2)y = 1/2AB ops in 60 = 1.5x
3)① If the PA bisector ∠BPO, the distance from A to BP is H = 2 ∠ 3 according to the nature of the angular bisector.
Equal product method of △BAP: 1.5x = 1/2bp 2 √ 3.
∴BP=√3x/2, that is, PB⊥BO at this time.
∴x=2OB=8+4√3
∠BPA= 1/2∠BPO= 15
② Existence, x=2+√3