解:(1)∵Rt△ABO,點(diǎn)B在y軸上,且OB=4,sinA=
,
∴sinA=
=
,
∴AO=5,AB=3,
∴點(diǎn)A的坐標(biāo)為:(3,4),
∴反比例函數(shù)y=
(x>0)的解析式為:xy=12,
∴y=
;
(2)∵將△ABO沿直線y=x翻折,折疊后點(diǎn)B的對(duì)應(yīng)點(diǎn)為B′,點(diǎn)A的對(duì)應(yīng)點(diǎn)為A′,
∴BO=B′O,
∴B′點(diǎn)的坐標(biāo)為:(4,0),
∵A′B′=AB=3,
∴A′點(diǎn)的坐標(biāo)為:(4,3),
∵4×3=12,
∴點(diǎn)A′落在反比例函數(shù)y=
(x>0)的圖象上.
分析:(1)根據(jù)OB=4,sinA=
,即可得出AO,以及AB的長(zhǎng),即可得出A點(diǎn)坐標(biāo)以及反比例函數(shù)y=
(x>0)的解析式.
(2)根據(jù)將△ABO沿直線y=x翻折,折疊后點(diǎn)B的對(duì)應(yīng)點(diǎn)為B′,點(diǎn)A的對(duì)應(yīng)點(diǎn)為A′,利用對(duì)稱性質(zhì)求出B′點(diǎn)的坐標(biāo)即可,進(jìn)而將A′代入解析式,判斷出是否落在反比例函數(shù)y=
(x>0)的圖象上.
點(diǎn)評(píng):此題主要考查了解直角三角形的性質(zhì)以及反比例函數(shù)的性質(zhì),根據(jù)已知得出A點(diǎn)坐標(biāo)以及利用對(duì)稱性求出A′點(diǎn)的坐標(biāo)是解題關(guān)鍵.