※ 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)