(有圖)數學問題~有關求圓的半徑及座標!!!－aptzmcv的部落格

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(有圖)數學問題~有關求圓的半徑及座標!!!

發問:

已知座標(X,Y) A(450.000m,160.000m) B(410.400m,260.500m) 角度 a =49度 b =9度 r =10度距離AS1=48.870m AS2=44.170m AS3=45.110m 1.求圓的半徑 2.求圓中心點座標 (圖檔) http://home.pchome.com.tw/net/bluejam318/PIC/321.JPG

最佳解答:

向量AB = (410.4-450,260.5-160) = (-39.6,100.5) 假設向量AS1 = (x1-450,y1-160) = (X1,Y1) 向量AS2 = (x2-450,y2-160) = (X2,Y2) 向量AS3 = (x3-450,y3-160) = (X3,Y3) 向量長度 |a| 純量積公式向量a?向量b = |a|*|b|*cosθ AS1 = 48.87 = √(X1^2+Y1^2) (X1,Y1)?(-39.6,100.5) = √(X1^2+Y1^2)*√[(-39.6)^2+100.5^2)]*cos(49°) => X1^2+Y1^2 = (48.87)^2...(1) -39.6*X1+100.5*Y1 = √(X1^2+Y1^2)*108.0204*0.6561...(2) (1)代入(2) => X1^2+Y1^2 = 2388.2769 -39.6*X1+100.5*Y1 = 48.87*108.0204*0.6561 => X1^2+Y1^2 = 2388.2769...(3) Y1 = 0.394*X1+34.4629...(4) (4)代入(3) => X1^2+(0.394*X1+34.4629)^2 = 2388.2769 => 1.1552*X1^2+27.1568*X1-1200.5854 = 0 => X1 = 22.5598 or -46.0681 (取正) => Y1 = 0.394*22.5598+34.4629 = 43.3515 向量AS1 = (22.5598,43.3515) = (x1-450,y1-160) => 點S1 = (472.5598,203.3515) AS2 = 44.17 = √(X2^2+Y2^2) (X2,Y2)?(-39.6,100.5) = √(X2^2+Y2^2)*√[(-39.6)^2+100.5^2)]*cos(49°+9°) => X2^2+Y2^2 = (44.17)^2...(5) -39.6*X2+100.5*Y2 = √(X2^2+Y2^2)*108.0204*0.5299...(6) (5)代入(6) => X2^2+Y2^2 = 1950.9889 -39.6*X2+100.5*Y2 = 44.17*108.0204*0.5299 => X2^2+Y2^2 = 1950.9889...(7) Y2 = 0.394*X2+25.1571...(8) (8)代入(7) => X2^2+(0.394*X2+25.1571)^2 = 1950.9889 => 1.1552*X2^2+19.8238*X2-1318.1092 = 0 => X2 = 26.2715 or -43.432 (取正) => Y2 = 0.394*26.2715+25.1571 = 35.5081 向量AS2 = (26.2715,35.5081) = (x2-450,y2-160) => 點S2 = (476.2715,195.5081) AS3 = 45.11 = √(X3^2+Y3^2) (X3,Y3)?(-39.6,100.5) = √(X3^2+Y3^2)*√[(-39.6)^2+100.5^2)]*cos(49°+9°+10°) => X3^2+Y3^2 = (45.11)^2...(9) -39.6*X3+100.5*Y3 = √(X3^2+Y3^2)*108.0204*0.3746...(10) (9)代入(10) => X3^2+Y3^2 = 2034.9121 -39.6*X3+100.5*Y3 = 45.11*108.0204*0.3746 => X3^2+Y3^2 = 2034.9121...(11) Y3 = 0.394*X3+18.1627...(12) (12)代入(11) => X3^2+(0.394*X3+18.1627)^2 = 2034.9121 => 1.1552*X3^2+14.3122*X2-1705.0284 = 0 => X3 = 32.7198 or -45.1091 (取正) => Y3 = 0.394*32.7198+18.1627 = 31.0543 向量AS3 = (32.7198,31.0543) = (x3-450,y3-160) => 點S3 = (482.7198,191.0543) 假設圓心O為(X0,Y0) 則線段OS1 = 線段OS2 = 線段OS3 =>√[(X0-472.5598)^2+(Y0-203.3515)^2] = √[(X0-476.2715)^2+(Y0-195.5081)^2] √[(X0-472.5598)^2+(Y0-203.3515)^2] = √[(X0-482.7198)^2+(Y0-191.0543)^2] √[(X0-476.2715)^2+(Y0-195.5081)^2] = √[(X0-482.7198)^2+(Y0-191.0543)^2] =>(X0-472.5598)^2+(Y0-203.3515)^2 = (X0-476.2715)^2+(Y0-195.5081)^2 (X0-472.5598)^2+(Y0-203.3515)^2 = (X0-482.7198)^2+(Y0-191.0543)^2 (X0-476.2715)^2+(Y0-195.5081)^2 = (X0-482.7198)^2+(Y0-191.0543)^2 =>7.4234*X0-15.6868*Y0 = 393.3617 20.3200*X0-24.5944*Y0 = 4855.5537 12.8966*X0-8.9076*Y0 = 4462.1920 =>X0 = 488.2712 , Y0 = 205.9866 則圓心O為(488.2712,205.9866) 圓的半徑為 √[(488.2712-472.5598)^2+(205.9866-203.3515)^2] = 15.9308 m 2006-03-10 13:27:48 補充：對不起，我沒有在用MSN和YAHOO MSN，你可以在意見欄問我，我再跟你討論。(對不起我本來寫的更多，可是有2000字的限制，所以刪掉了一些過程。 2006-03-10 23:03:10 補充：嗯!可以用代數解聯立方程式，不過還是要用三角函數建立關係式，那就跟像量差不多了。我寄文字檔給你好了^_^

其他解答:

挖賽.......盡然有著ㄇ複雜ㄉ數學題目...........|||||嗯嗯~那這一題只能用向量解囉?? 因為我是學測量的!! 2006-03-10 23:15:22 補充：嗯嗯~謝謝你喔請問你讀哪阿~幾年級呢?? 真厲害喔~~感謝!!481517FD598DAD6B