题目内容
(2002•安徽)附加题:如图,在△ABC中,AB=5,AC=7,∠B=60°,求BC的长.(华东版教材实验区试题)![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_ST/images0.png)
【答案】分析:可通过构建直角三角形来求解,过点A作AD⊥BC于D,AD是公共直角边,因此先求出AD是解题的关键,在Rt△ABD中,有AB的长,有∠B的度数,可以求出BD的长,AD的长,在Rt△ADC中,求出了AD的长,有AC的长,因此根据勾股定理可求出CD的长,有了BD、CD的长,也就求出了BC的长.
解答:
解:过点A作AD⊥BC于D,
在Rt△ABD中,
AD=AB•sin60°=5×
=
,
BD=AB•cos60°=5×
=
,
在Rt△ADC中,
DC=
=
=
,
所以BC=BD+DC=8.
点评:在用解直角三角形的方法求线段长的时候,没有直角三角形的条件下,要根据已知条件构建直角三角形进行求解.
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/images0.png)
在Rt△ABD中,
AD=AB•sin60°=5×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/1.png)
BD=AB•cos60°=5×
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/2.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/3.png)
在Rt△ADC中,
DC=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/5.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/20131019105453850806075/SYS201310191054538508060025_DA/6.png)
所以BC=BD+DC=8.
点评:在用解直角三角形的方法求线段长的时候,没有直角三角形的条件下,要根据已知条件构建直角三角形进行求解.
![](http://thumb2018.1010pic.com/images/loading.gif)
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