As shown in the figure, it is known that AB is the diameter of ⊙ o, PA and PC are the tangents of ⊙ o, a and C are the tangent points, ∠ BAC = 30 ° (I) find the size of ∠ p; (II) if AB = 2, find the length of PA (the result retains the root sign)

(I) ∵ PA is the tangent of ⊙ o, AB is the diameter of ⊙ o, ∵ PA ⊥ AB, ∵ BAP = 90 °; ∵ BAC = 30 °, ∵ cap = 90 ° - ∵ BAC = 60 °. PA and PC cut ⊙ o at points a and C, ∵ PA = PC, ∵ PAC is an equilateral triangle, ∵ P = 60 ° (II) as shown in the figure, connect BC, then ∵ ACB = 90 °. In RT △ ACB, ab = 2, ∵ BAC = 30 °, ∵ cos ∵ BAC = acab, ∵ AC = ab · cos ∵ BAC = 2cos 30 ° = 3. ∵ △ PAC is an equilateral triangle, ∵ PA = AC, ∵ PA = 3

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