【答案】
分析:(1)先由直線AB的解析式求A、B兩點的坐標,再根據(jù)銳角三角函數(shù)值求∠ABO的度數(shù);
(2)由∠EBP=∠ABO,已知BP,解直角三角形EBP求BE;
(3)①過點C作CM⊥OA于M,在直角三角形ACM中,已知AC及∠CAM的度數(shù),根據(jù)銳角三角函數(shù)即可求出點C的坐標;
②要使△EPC和△AOB相似,而△AOB是有一個角為30°的直角三角形,只需△EPC也是有一個角為30°的直角三角形.由于∠CEP<∠BEP=90°,所以有可能∠CPE=90°或者∠PCE=90°,然后分情況討論.
解答:解:(1)∵直線AB:
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/0.png)
與y軸、x軸交于A、B兩點,
∴A(0,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/1.png)
),B(1,0).
在直角△AOB中,∵tan∠ABO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/3.png)
,
∴∠ABO=60°;
(2)當t=5時,BP=4,
在直角△EBP中,∠BEP=90°,∠EBP=∠ABO=60°,
∴BE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/4.png)
BP=2;
(3)①過點C作CM⊥OA于M.
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/images5.png)
∵將△AOB沿直線AB翻折180°,得到△ABC,
∴△AOB≌△ACB,
∴∠OAB=∠CAB=30°,AO=AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/5.png)
,
∴∠MAC=60°.
在直角三角形ACM中,∠AMC=90°,AC=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/6.png)
,∠CAM=60°,
∴CM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/7.png)
,AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/8.png)
,
∴OM=OA-AM=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/9.png)
.
∴點C的坐標為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/10.png)
,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/11.png)
);
②∵△EPC和△AOB相似,∠CEP<∠BEP=90°,
∴可能∠CPE=90°或者∠PCE=90°,且△EPC有一個角為30°.
設E(x,-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/12.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/13.png)
),點P的坐標為(t,0).
過點E作EN⊥OP于N,由射影定理,得EN
2=BN•NP,
即(-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/14.png)
x+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/15.png)
)
2=(x-1)(t-x),
整理,得t=4x-3.
分如下幾種情況:
第一種:如果∠CPE=90°,∠CEP=30°,那么CP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/16.png)
CE,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/images18.png)
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/17.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/18.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/19.png)
,
整理,得20x
2-46x+27=0,
∵△=(-46)
2-4×20×27<0,
∴原方程無解;
第二種:如果∠CPE=90°,∠ECP=30°,那么EP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/20.png)
CE,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/21.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/22.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/23.png)
,
整理,得44x
2-90x+45=0,
∵△=(-90)
2-4×44×45=180,
∴x=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/24.png)
,
∴t=4x-3=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/25.png)
,
又∵t>1,
∴t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/26.png)
;
第三種:如果∠PCE=90°,∠CEP=30°,那么CP=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/27.png)
PE,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/28.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/29.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/30.png)
,
整理,得13x
2-30x+18=0,
∵△=(-30)
2-4×13×18<0,
∴原方程無解;
第四種:如果∠PCE=90°,∠CPE=30°,那么CE=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/31.png)
PE,
即
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/32.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/33.png)
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/34.png)
,
整理,得x
2=0,
∴x=0,
∴t=4x-3=-3,不合題意舍去,
∴原方程無解.
綜上,可知當t=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131022163650432733963/SYS201310221636504327339023_DA/35.png)
時,△EPC和△AOB相似.
點評:本題主要考查了一次函數(shù),直角三角形、全等三角形、相似三角形的知識,綜合性強,有一定難度.運用分類討論的思想解決最后一問是解題的關鍵.