※ 引述《EIORU ()》之铭言： : 一个25m深水井, : | | : d| | : | |a : | e | : | | : b一一一c : 斜放了两个长度为20m(cd相连线段)和15m(ab相连线段)的梯子 : 若此两个梯子在e的地方交叉 : 已知e点距离井口17m : 求此井的宽度(bc线段长), 以及角aed的大小 由题意知e点离井底25-17=8m 由e向井底作垂直 垂足为h 则 de:ec = bh:hc = be:ea (由db//eh//ac知) 设井宽x 上述相等的比de/ec = bh/hc = be/ea = r 则 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长 再由三角形ebc可求出角bec 即角aed之大小 -- 详细数字等我回去开软体来算= = 手解四次方程很累人的... -- 回来开了软体来算 上面的四次式 有两实根两虚根 实根一正一负 显然这里的r只能是正的 所以仅有唯一合理解 约为1.38675 代回得x约为5.95132 即为bc长 然後eb为ab的r/(r+1) = 15 * r/(r+1) = 8.71531 ec为cd的1/(r+1) = 20 * 1/(r+1) = 8.37958 由余弦定理得 cos 角bec = (eb^2 + ec^2 - bc^2)/2*eb*ec = 0.758283 得该角约为 0.71 弪 ≒ 40.68 度 -- 你是不是要写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 --

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