题目内容
如图△ABC中,
为直角,
于
,
, DB = , CD = ![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230125049131598.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504835383.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504851569.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504867323.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504898684.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230125049131598.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504929389.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504945420.png)
分析:由△ABC中,∠C为直角,CD⊥AB,根据等角的余角相等,即可求得∠BCD=∠A,又由BC=3,AB=5,利用勾股定理即可求得AC的长,然后在Rt△BCD中,利用三角函数的知识即可求得答案.
解:∵CD⊥AB,
∴∠CDB=90°,
∴∠B+∠BCD=90°,
∵△ABC中,∠C为直角,
∴∠A+∠B=90°,
∴∠A=∠BCD,
∵BC=3,AB=5,
∴AC=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504976671.png)
在Rt△BCD中,DB=BC?sin∠BCD=BC?sin∠A=3×
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012504991369.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012505007383.png)
CD=BC?cos∠BCD=BC?cos∠A=3×
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012505023346.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012505038412.png)
故答案为:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012505007383.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823012505038412.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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