【答案】
分析:(1)連OH,作HK⊥x軸于k,根據(jù)關(guān)于x軸對稱的坐標(biāo)特點(diǎn)得到H點(diǎn)坐標(biāo)為(3,-4),再根據(jù)切線的性質(zhì)由AH為⊙O的切線,得到OH⊥AH,利用等角的余角相等得到∠OAH=∠KHO,根據(jù)三角形相似的判定得RtAKH∽Rt△HKO,則AK:HK=HK:OK,即AK:4=4:3,求出AK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/0.png)
,易得A點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/1.png)
,0),然后利用待定系數(shù)法求直線AH的解析式;
(2)在Rt△OKH中,利用勾股定理計算出OH=5,然后在Rt△OAH中,利用正弦的定義即可得到sin∠HAO的值;
(3)過點(diǎn)D作DM⊥EF于M,并延長DM交⊙O于N,連接ON,交BC于T,根據(jù)垂徑定理得到OM垂直平分DN,即D點(diǎn)與N點(diǎn)關(guān)于x軸對稱,則N點(diǎn)坐標(biāo)為(3,-4),ON=5;由DM⊥EF根據(jù)等腰三角形的性質(zhì)可得DN平分∠BDC,即∠CDN=∠BDN,根據(jù)圓周角定理的推論得到弧BN=弧CN,然后利用垂徑定理的推論可得OT⊥BC,利用等角的余角相等得到∠TGO=∠MNO,在Rt△OMN,OM=3,MN=4,利用正弦的定義即可得到sin∠MNO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/2.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/3.png)
,則sin∠CGO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/4.png)
,即sin∠CGO的大小不變.
解答:解:(1)如圖,連OH,作HK⊥x軸于k,
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/images5.png)
∵點(diǎn)D(3,4),點(diǎn)H與點(diǎn)D關(guān)于x軸對稱,
∴H點(diǎn)坐標(biāo)為(3,-4),
∵AH為⊙O的切線,
∴OH⊥AH,
∴∠AOH+∠OAH=90°,∠KOH+∠KHO=90°,
∴∠OAH=∠KHO,
∴Rt△AKH∽Rt△HKO,
∴AK:HK=HK:OK,即AK:4=4:3,
∴AK=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/5.png)
,
∴OA=OK+AK=3+
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/6.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/7.png)
,
∴A點(diǎn)坐標(biāo)為(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/8.png)
,0),
設(shè)直線HA的函數(shù)解析式為y=kx+b,
把H(3,-4),A(
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/9.png)
,0)代入得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/10.png)
,
解得
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/11.png)
,
∴直線HA的函數(shù)解析式為y=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/12.png)
x-
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/13.png)
;
(2)在Rt△OKH中,OH=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/14.png)
=5,
在Rt△OAH中,sin∠HAO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/15.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/16.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/17.png)
;
(3)sin∠CGO的大小不變.理由如下:
過點(diǎn)D作DM⊥EF于M,并延長DM交⊙O于N,連接ON,交BC于T,如圖,
則OM垂直平分DN,即D點(diǎn)與N點(diǎn)關(guān)于x軸對稱,
則N點(diǎn)坐標(biāo)為(3,-4),ON=5,
又∵△DEF為等腰三角形,DM⊥EF,
∴DN平分∠BDC,即∠CDN=∠BDN,
∴弧BN=弧CN,
∴OT⊥BC,
∴∠TGO+∠GOT=90°,
而∠MNO+∠MON=90°,
∴∠TGO=∠MNO,
在Rt△OMN,OM=3,MN=4,
∴sin∠MNO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/18.png)
=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/19.png)
,
∴sin∠CGO=
![](http://thumb.zyjl.cn/pic6/res/czsx/web/STSource/20131101185545823520839/SYS201311011855458235208022_DA/20.png)
.
即當(dāng)E、F兩點(diǎn)在OP上運(yùn)動時(與點(diǎn)P不重合),sin∠CGO的值不變.
點(diǎn)評:本題考查了圓的綜合題:圓的切線垂直于過切點(diǎn)的半徑;垂直于弦的直徑平分弦,并且平分弦所對的��;平分弦所對的弧的直徑垂直平分弦;在同圓或等圓中,相等的圓周角所對的弧相等;運(yùn)用待定系數(shù)法求函數(shù)的解析式以及關(guān)于坐標(biāo)軸對稱的點(diǎn)的坐標(biāo)特點(diǎn);運(yùn)用相似三角形的判定與性質(zhì)和勾股定理進(jìn)行幾何計算.