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請問為何AB線段長=2Rsin(theta/2)
三角形的邊角關係
2013-05-29 8:15 pm
回答 (2)
2013-05-29 9:36 pm
2013-05-29 9:16 pm
從圓心O畫OH垂直AB線段(弦)交於H點
∠AOH=θ/2，AB=2AH
依三角函數定義，
sin(θ/2) = AH/R => AH=Rsin(θ/2)
所以，AB=2AH=2Rsin(θ/2)
∠AOH=θ/2，AB=2AH
依三角函數定義，
sin(θ/2) = AH/R => AH=Rsin(θ/2)
所以，AB=2AH=2Rsin(θ/2)
收錄日期: 2021-05-01 18:57:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130529000015KK01651