【答案】
分析:(1)過點D作DA⊥OB,垂足為A.利用三角函數(shù)可求得,點D的坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/0.png)
),設(shè)直線DB的函數(shù)表達(dá)式為y=kx+b,把點B(5,0),D(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/1.png)
)代入解析式利用待定系數(shù)法,得直線DB的函數(shù)表達(dá)式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/2.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/3.png)
;
(2)先證明△ODM∽△BMC.得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/4.png)
,所以O(shè)D•BC=BM•OM.設(shè)OM=x,則BM=5-x,得2×2=x(5-x),解得x的值,即可求得M點坐標(biāo);
(3)(Ⅰ)當(dāng)M點坐標(biāo)為(1,0)時,如圖①,OM=1,BM=4.先求得DME∽△CMF,所以
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/5.png)
,
可得CF=2DE.所以2-n=2m,即m=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/6.png)
.(Ⅱ)當(dāng)M點坐標(biāo)為(4,0)時,OM=4,BM=1.同(Ⅰ),可得△DME∽△CMF,得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/7.png)
,所以DE=2CF.解得m=2(2-n),即m=4-2n.
解答:![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/images8.png)
解:(1)過點D作DA⊥OB,垂足為A.
在Rt△ODA中,∠DAO=90°,∠DOB=60°,
∴DA=OD•sin∠DOB=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/8.png)
,
OA=OD•cos∠DOB=1,
∴點D的坐標(biāo)為(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/9.png)
),
設(shè)直線DB的函數(shù)表達(dá)式為y=kx+b,
由B(5,0),D(1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/10.png)
),得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/11.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/12.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/images14.png)
∴直線DB的函數(shù)表達(dá)式為y=-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/13.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/14.png)
;
(2)∵∠DMC=∠DOB=60°,
∴∠ODM+∠DMO=120°,∠DMO+∠CMB=120°,
∴∠ODM=∠CMB,
∵等腰梯形ABCD的∠DOB=∠CBO,
∴△ODM∽△BMC,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/15.png)
,
∴OD•BC=BM•OM,
∵B點為(5,0),
∴OB=5.
設(shè)OM=x,則BM=5-x
∵OD=BC=2,
∴2×2=x(5-x),
解得x
1=1,x
2=4,
∴M點坐標(biāo)為(1,0)或(4,0);
(3)解:(Ⅰ)當(dāng)M點坐標(biāo)為(1,0)時,如圖1,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/images18.png)
OM=1,BM=4.
∵DC∥OB,
∴∠MDE=∠DMO,
又∵∠DMO=∠MCB,
∴∠MDE=∠MCB,
∵∠DME=∠CMF=α,
∴△DME∽△CMF,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/16.png)
,
∴CF=2DE,
∵CF=2-n,DE=m,
∴2-n=2m,即m=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/17.png)
;
(Ⅱ)當(dāng)M點坐標(biāo)為(4,0)時,如圖2
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/images21.png)
OM=4,BM=1.
同(Ⅰ),可得△DME∽△CMF,
∴
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/18.png)
,
∴DE=2CF,
∵CF=2-n,DE=m,
∴m=2(2-n),即m=4-2n.
綜上所述,m與n的函數(shù)關(guān)系式為:m=1-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131021231904443565016/SYS201310212319044435650010_DA/19.png)
或m=4-2n.
點評:主要考查了函數(shù)和幾何圖形的綜合運用,其涉及的知識點比較多.解題的關(guān)鍵是會靈活的運用函數(shù)圖象的性質(zhì)和交點的意義結(jié)合梯形的性質(zhì)利用相似比中的成比例線段作為相等關(guān)系求線段之間的等量關(guān)系.試題中貫穿了方程思想和數(shù)形結(jié)合的思想,請注意體會.