题目内容
(2002•内江)已知如图,在Rt△ABC中,∠C=90°,D是BC中点,DE⊥AB,垂足为E,∠B=30°,AE=7.求:DE的长.![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_ST/images0.png)
【答案】分析:应设所求的线段为未知数,用三角函数表示出AB、BE,进而表示出DE,求解即可.
解答:解:设DE=x,
∵DE⊥AB,∠B=30°,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/images0.png)
∴BE=
x,BD=2x,
∵D是BC中点,
∴BC=4x,
在Rt△ABC中,可得到:AB=
=
=
x,
∵AB-BE=7,
∴
x-
x=7,
解得x=
.
点评:解决本题的关键是利用三角函数表示出题中唯一给出的线段的长.
解答:解:设DE=x,
∵DE⊥AB,∠B=30°,
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/images0.png)
∴BE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/0.png)
∵D是BC中点,
∴BC=4x,
在Rt△ABC中,可得到:AB=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/1.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/3.png)
∵AB-BE=7,
∴
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/5.png)
解得x=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131021233025649429149/SYS201310212330256494291024_DA/6.png)
点评:解决本题的关键是利用三角函数表示出题中唯一给出的线段的长.
![](http://thumb2018.1010pic.com/images/loading.gif)
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