如图,平行四边形ABCD中,AB=8cm,BC=6cm,∠A=30°,点P从点A沿线段AB以1cm/s的速度向

如图，平行四边形ABCD中，AB=8cm，BC=6cm，∠A=30°，点P从点A沿线段AB以1cm/s的速度向点B移动.
1）当P点运动几秒时，△PBC为等腰三角形；（2）设S△PBC=y，请写出y（cm2）与点P的移动时间 t（s）之间的函数关系式，并写出t的取值范围；（3）是否存在否一点P，使S△PBC=1/3S平行四边形ABCD？若存在求AP的长，若不存在，请说明理由。

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(1) As the angle B is greater than 90 degrees, the only possibility will be BP=BC.
BC=6, AB=8, hence AP=2. t=2/1=2(second)
(2) We can see that AP=t*1=t, then BP=8-t,
(the height of PBC on the BP side)=BC*sinA=6*sin30=3
y=(8-t)*3/2=12-3/2t
(3) Yes, there is such a point P.
Area of ABCD=8*3=24
Area of PBC should be: 24/3=8
That is: 12-3/2t=8, t=8/3
AP=8/3
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