题目内容
已知:如图所示,Rt△ABC中,∠C=90°,∠ABC=60°,DC=11,D点到AB的距离为2,求BD的长.![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_ST/images0.png)
【答案】分析:根据DE求AD的长度,根据AC=AD+CD即可求AC的长度,∵∠A=30°,∴AE=
DE=2
,且根据AB=2BC,AB2-BC2=AC2即可求得AB的长度,∴BE=AB-AE,根据BD=
即可求BD的长.
解答:解:DE⊥AB于E,且DE=2,
在Rt△ABC中,
∵∠ABC=60°,
∴∠A=30°.
在Rt△ADE中,
∵DE=2,
∴AD=4,AE=
,
∵DC=11,∴AC=11+4=15,
∴AB=![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/4.png)
∴
,
在Rt△DEB中,
,
∴BD=14.
答:BD的长为14.
点评:本题考查了勾股定理的运用,考查了30°角在直角三角形中的运用,本题中巧妙地利用∠A=30°是解题的关键.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/2.png)
解答:解:DE⊥AB于E,且DE=2,
在Rt△ABC中,
∵∠ABC=60°,
∴∠A=30°.
在Rt△ADE中,
∵DE=2,
∴AD=4,AE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/3.png)
∵DC=11,∴AC=11+4=15,
∴AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/4.png)
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/5.png)
在Rt△DEB中,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233151474798711/SYS201310212331514747987017_DA/6.png)
∴BD=14.
答:BD的长为14.
点评:本题考查了勾股定理的运用,考查了30°角在直角三角形中的运用,本题中巧妙地利用∠A=30°是解题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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