题目内容
如图PAB、PCD是⊙O的两条割线,AB是⊙O的直径.(1)如图甲,若PA=8,PC=10,CD=6.
①求sin∠APC的值;②sin∠BOD=______
【答案】分析:(1)作OE⊥CD于E,连接OC,作DF⊥PB于F.根据垂径定理和勾股定理求得有关线段的长,再进一步根据锐角三角函数的概念求解;
(2)根据圆周角定理的推论和等弧对等弦进行证明,根据平行线等分线段定理进行求解.
解答:解:(1)作OE⊥CD于E,连接OC,作DF⊥PB于F.
①根据垂径定理,得CE=3.设圆的半径是r.
根据勾股定理,得![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/images0.png)
OP2-PE2=OC2-CE2,
(8+r)2-169=r2-9,
解得r=6.
则OE=3
.
则sin∠APC=
=
;
②设OF=x.
根据勾股定理,得
PD2-PF2=OD2-OF2,
256-(14+x)2=36-x2,
解得x=
.
所以DF=
.
所以sin∠BOD=
=
.
(2)①∵AC∥OD,
∴∠1=∠2.
又OA=OD,
∴∠2=∠3.
∴∠1=∠3.
所以弧CD=弧BD,
所以CD=BD;
②∵AC∥OD,
∴
=
.
又CD=BD,AB=2OA,
∴
=
.
∴cos∠BAD=
=
.
点评:此题综合运用了垂径定理、勾股定理、圆周角定理的推论、等弧对等弦、平行线等分线段定理等,有一定的难度.
(2)根据圆周角定理的推论和等弧对等弦进行证明,根据平行线等分线段定理进行求解.
解答:解:(1)作OE⊥CD于E,连接OC,作DF⊥PB于F.
①根据垂径定理,得CE=3.设圆的半径是r.
根据勾股定理,得
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/images0.png)
OP2-PE2=OC2-CE2,
(8+r)2-169=r2-9,
解得r=6.
则OE=3
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/0.png)
则sin∠APC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/2.png)
②设OF=x.
根据勾股定理,得
PD2-PF2=OD2-OF2,
256-(14+x)2=36-x2,
解得x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/3.png)
所以DF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/4.png)
所以sin∠BOD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/6.png)
(2)①∵AC∥OD,
∴∠1=∠2.
又OA=OD,
∴∠2=∠3.
∴∠1=∠3.
所以弧CD=弧BD,
所以CD=BD;
②∵AC∥OD,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/7.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/8.png)
又CD=BD,AB=2OA,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/9.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/10.png)
∴cos∠BAD=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/11.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131103000437003307185/SYS201311030004370033071026_DA/12.png)
点评:此题综合运用了垂径定理、勾股定理、圆周角定理的推论、等弧对等弦、平行线等分线段定理等,有一定的难度.
![](http://thumb2018.1010pic.com/images/loading.gif)
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