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806 B
806 B
根据条件，知道OA=\sqrt{3},OB=1
\therefore AB=2,\angle OAB=30^{\circ},\angle OBA=60^{\circ}
第一问
当A'B \perp OB时,A'B=\sqrt{2},OA'=OA=\sqrt{3}
\therefore OB=\sqrt{\sqrt{3}^2=\sqrt{2}^2}=1
\therefore A'(\sqrt{2},1)
第二问
P是中点，则OP是直角三角形AOB的斜边中线，
\therefore OP=AP=A'P=BP=1
\because \angle OBA=60^{\circ}
\therefore \triangle OPB是等边三角形
\therefore \angle OPB=60^{\circ}
\therefore \angle OPA=120^{\circ}
由于翻折，所以\angle OPA=\angle OPA'
\therefore \angle BPA'=60^{\circ}
\therefore \triangle A'BP是等边三角形
\therefore A'B=1