题目内容
(2009•大兴区二模)如图,电线杆AB直立于地面上,它的影子恰好照在土坡的坡面CD和地面BC上,若CD与地面成45°,∠A=60°,CD=4m,![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_ST/images1.png)
【答案】分析:延长AD交地面于E,作DF⊥BE于F,求出BE=BC+CF+FE=
,根据正切求出AB的值即可.
解答:
解:延长AD交地面于E,作DF⊥BE于F.
∵∠DCF=45°.CD=4.
∴CF=DF=
.
由题意知AB⊥BC.
∴∠EDF=∠A=60°.
∴∠DEF=30°
∴EF=
.
∴BE=BC+CF+FE=
.
在Rt△ABE中,∠E=30°.
∴AB=BEtan30°=
(m).
答:电线杆AB的长为6
米.
点评:此题主要是运用所学的解直角三角形的知识解决实际生活中的问题.作辅助线、求出BE=BC+CF+FE是解题的关键.
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/0.png)
解答:
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/images1.png)
∵∠DCF=45°.CD=4.
∴CF=DF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/1.png)
由题意知AB⊥BC.
∴∠EDF=∠A=60°.
∴∠DEF=30°
∴EF=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/2.png)
∴BE=BC+CF+FE=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/3.png)
在Rt△ABE中,∠E=30°.
∴AB=BEtan30°=
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/4.png)
答:电线杆AB的长为6
![](http://thumb.1010pic.com/pic6/res/czsx/web/STSource/201310201205326357056032/SYS201310201205326357056016_DA/5.png)
点评:此题主要是运用所学的解直角三角形的知识解决实际生活中的问题.作辅助线、求出BE=BC+CF+FE是解题的关键.
![](http://thumb2018.1010pic.com/images/loading.gif)
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