试题分析:(1)连接OQ,由PN切⊙O于Q,根据切线的性质可得OQ⊥PN,又由PO=20cm,∠P=30°,即可求得PQ的长; (2)作OE⊥BA于E,由BA⊥PN,即可得四边形AHOQ是矩形,当矩形AEOQ是正方形时,直线BA与⊙O相切.即可求得PB与BA的长,然后分别从当PQ﹣PA=OQ时,直线BA第一次与⊙O相切与当PA﹣PQ=OQ时,直线BA第二次与⊙O相切去分析求解,即可求得答案. 试题解析:(1)解:连结OQ,如图1
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105005849-27087.png) ∵PN与⊙O相切于点Q,∴OQ⊥PN,∵∠P=30°,OP=20,∴OQ=10,在Rt△OPQ中,
; (2)解:设运动t秒,BP=4t,则AB= ,AP= , ①如图2,当AB与⊙O切于点E时,连结OE,
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105005850-33192.png) ∴OE⊥AB,又∵OQ⊥PN,AB⊥PN,∴四边形AEOQ是矩形, ∴OE=AQ=10,∴ ,∴ , ②如图3,当A′B′与⊙O相切于点F时,连结OF, ∴OF⊥A′B′,又∵OQ⊥PN,AB⊥PN,∴四边形A′FOQ是矩形,∴OF=A′Q,∴ , ∴ ,∴当t为 秒或 秒时,直线AB与⊙O相切.
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