如图,已知AB是圆o的直径,弦CD交AB于E点,BE=1,AE=5,角AEC=30度,求cD的长
如题所述
第1个回答 2014-10-11
解:过O作OF⊥CD
∵AE=1,BE=5
∴AB=AE+BE=1+5=6
∴AO=AB/2=6/2=3
∴OE=AO-AE=3-1=2
∵∠AEC=45
∴∠OEF=45
∵OF⊥CD
∴OF=OE×√2/2=√2
∴CF=√(OC²-OF²)=√(9-2)=√7
∴CD=2CF=2√7
追答
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/0bd162d9f2d3572c615b1daa8913632762d0c30f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
追问为什么角BED为60度?应该是30度吧
有点不懂,你可以稍微解释一下吗?
追答恩。等一下
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/f11f3a292df5e0fe412601775f6034a85edf728e?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
现在明白了?
追问谢谢
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