题目内容
如图,AB为⊙O的直径,点C在⊙O上,过点C作⊙O的切线交AB的延长线于点D,已知∠D=30°
小题1:求∠A的度数;
小题2:若点F在⊙O上,CF⊥AB,垂足为E,CF=
,求图中阴影部分的面积.![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135310211640.png)
小题1:求∠A的度数;
小题2:若点F在⊙O上,CF⊥AB,垂足为E,CF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531005415.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135310211640.png)
小题1:30°
小题2:
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531036411.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531052435.png)
解:(1)连接OC
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135310672402.png)
∵CD切⊙O点C∴∠OCD=90°
又∵∠D=30°∴∠COD=60°又∵OA=OC∴∠A=∠ACO=30°
(2)由题意得∠OCF=30°,CF=
CE=
,
∴OC=4故![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135311141311.png)
∴
π![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531052435.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135310672402.png)
∵CD切⊙O点C∴∠OCD=90°
又∵∠D=30°∴∠COD=60°又∵OA=OC∴∠A=∠ACO=30°
(2)由题意得∠OCF=30°,CF=
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531005415.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531099423.png)
∴OC=4故
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135311141311.png)
∴
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/201408230135312551022.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531036411.png)
![](http://thumb.zyjl.cn/pic2/upload/papers/20140823/20140823013531052435.png)
![](http://thumb.zyjl.cn/images/loading.gif)
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