Re: [幾何] 角度問題
※ 引述《a5718090610 (Linear)》之銘言：
: When lengths of AB, BC and CD are equal, find the angle ∠BCD
: https://www.dropbox.com/s/h24ezrur327wjo0/%E6%9C%AA%E5%91%BD%E5%90%8D.png
: 這個是圖~
: 不好意思~小弟解了1H都解不出來~請問有人可以幫我解出來嗎? 謝謝~
設AB = a
∠CBD = 90 - x/2
以下角度均用度數
BD = 2asin(x/2)
2asin(x/2) a
---------- = -------------
sin(54) sin(108 - x/2)
=> 2sin(x/2)[sin(108)cos(x/2) - cos(108)sin(x/2)] = sin(54)
=> sin(108)sin(x) + cos(108)cos(x) = sin(54) + cos(108)
=> cos(108 - x) = sin(54) + cos(108)
=> x = 108 - arccos(sin(54) + cos(108)) 度數
--
※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 220.141.64.179
※ 文章網址: http://www.ptt.cc/bbs/Math/M.1400245394.A.451.html
→
05/16 22:20, , 1F
05/16 22:20, 1F
討論串 (同標題文章)
