Re: [問題] 數學題目

看板Inference (推理遊戲)作者LPH66 (ha(ruhi|yate)ism)時間17年前 (2007/05/10 18:45)推噓0(0推 0噓 0→)

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※ 引述《EIORU ()》之銘言： : 一個25m深水井, : | | : d| | : | |a : | e | : | | : b一一一c : 斜放了兩個長度為20m(cd相連線段)和15m(ab相連線段)的梯子 : 若此兩個梯子在e的地方交叉 : 已知e點距離井口17m : 求此井的寬度(bc線段長), 以及角aed的大小由題意知e點離井底25-17=8m 由e向井底作垂直垂足為h 則 de:ec = bh:hc = be:ea (由db//eh//ac知) 設井寬x 上述相等的比de/ec = bh/hc = be/ea = r 則 db:eh = bc:hc = (bh+hc):hc = (r+1):1 eh:ac = bh:bc = bh:(bh+hc) = 1:(1+1/r) = r:(r+1) 故eh = db/(r+1) = [r/(r+1)]*ac 又 db^2 = dc^2 - bc^2 = 20^2 - x^2 ac^2 = ab^2 - bc^2 = 15^2 - x^2 所以 20^2 - x^2 = [eh(r+1)]^2 15^2 - x^2 = [eh(r+1)/r]^2 20^2 - [eh(r+1)]^2 = 15^2 - [eh(r+1)/r]^2 代入eh=8 20^2 - [8(r+1)]^2 = 15^2 - [8(r+1)/r]^2 400 - 64(r^2+2r+1) = 225 - 64(1+2/r+1/r^2) 整理可得r的四次式由此解出r 代回上式可解出x 即bc長再由三角形ebc可求出角bec 即角aed之大小 -- 詳細數字等我回去開軟體來算= = 手解四次方程很累人的... -- 回來開了軟體來算上面的四次式有兩實根兩虛根實根一正一負顯然這裡的r只能是正的所以僅有唯一合理解約為1.38675 代回得x約為5.95132 即為bc長然後eb為ab的r/(r+1) = 15 * r/(r+1) = 8.71531 ec為cd的1/(r+1) = 20 * 1/(r+1) = 8.37958 由餘弦定理得 cos 角bec = (eb^2 + ec^2 - bc^2)/2*eb*ec = 0.758283 得該角約為 0.71 弳 ≒ 40.68 度 -- 你是不是要寫ACM的? -- 'Oh, Harry, dont't you see?' Hermione breathed. 'If she could have done one thing to make absolutely sure that every single person in this school will read your interview, it was banning it!' ---'Harry Potter and the order of the phoenix', P513 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 210.70.172.164 ※ 編輯: LPH66 來自: 192.192.197.116 (05/10 20:11)