![](http://thumb.1010pic.com/pic5/upload/201308/52846733e7ee1.png)
解:(1)如图所示,过点A作AE⊥BD于E,
设AB=a,
在Rt△ABC中,∠BCA=30°,那么可知
BC=cot30°×AB=
![](http://thumb.1010pic.com/pic5/latex/21.png)
a,
在Rt△BCD中,BD=sin45°×BC=
![](http://thumb.1010pic.com/pic5/latex/5054.png)
a,
又∵AE⊥BD,∠CBD=45°,
∴BE=AE=sin45°×a=
![](http://thumb.1010pic.com/pic5/latex/54.png)
a,
∴在Rt△ADE中,cot∠EDA=
![](http://thumb.1010pic.com/pic5/latex/182154.png)
=
![](http://thumb.1010pic.com/pic5/latex/479487.png)
=
![](http://thumb.1010pic.com/pic5/latex/479488.png)
=
![](http://thumb.1010pic.com/pic5/latex/21.png)
-1;
即cot∠BDA=
![](http://thumb.1010pic.com/pic5/latex/21.png)
-1.
(2)设所求作的二次项系数为1的一元二次方程为x
2+px+q=0,
∵cot∠BDA=
![](http://thumb.1010pic.com/pic5/latex/21.png)
-1,
∴tan∠BDA=
![](http://thumb.1010pic.com/pic5/latex/479489.png)
,
∴p=-(cot∠BDA+2tan∠BDA)=-(
![](http://thumb.1010pic.com/pic5/latex/21.png)
-1+2×
![](http://thumb.1010pic.com/pic5/latex/479490.png)
)=-2
![](http://thumb.1010pic.com/pic5/latex/21.png)
,
q=cot∠BDA×2tan∠BDA=(
![](http://thumb.1010pic.com/pic5/latex/21.png)
-1)×2×
![](http://thumb.1010pic.com/pic5/latex/190159.png)
=2,
∴所求作的一元二次方程为x
2-2
![](http://thumb.1010pic.com/pic5/latex/21.png)
x+2=0.
分析:(1)如图2先过点A作AE⊥BD于E,设AB=a,在Rt△ABC中,利用cot30°=
![](http://thumb.1010pic.com/pic5/latex/21.png)
,可求BC,在Rt△BCD中,利用sin45°=
![](http://thumb.1010pic.com/pic5/latex/54.png)
,又可求BD,易证△ABE是等腰直角三角形,从而利用sin45°=
![](http://thumb.1010pic.com/pic5/latex/54.png)
,可求AE、BE,于是在Rt△ADE中,可求cot∠EDA=
![](http://thumb.1010pic.com/pic5/latex/182154.png)
=
![](http://thumb.1010pic.com/pic5/latex/479487.png)
,即cot∠BDA的值.
(2)由(1)可求出tan∠BAD,设所求作的二次项系数为1的一元二次方程为x
2+px+q=0,根据根和系数的关系求出p和q,从得出一个二次项系数为1的一元二次方程,使cot∠BDA和2tan∠BDA为此方程的两个根.
点评:本题考查了直角三角形的性质、特殊三角函数值.解本题最关键的是作辅助线AE,构造直角三角形.