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”三角的應用”數學題一條 20fun x2
發問:
Two ships A and B leave a port at the same time.Ship A sails on a bearing of 300 degrees for 12km while ship B sails on a bearing of 150 degrees for 12km.Find:a)the distance between the two ships (answer is 23.2km)b)the bearing of ship A from ship B (answer is 315 degrees)c)the bearing of ship B from... 顯示更多 Two ships A and B leave a port at the same time. Ship A sails on a bearing of 300 degrees for 12km while ship B sails on a bearing of 150 degrees for 12km. Find: a)the distance between the two ships (answer is 23.2km) b)the bearing of ship A from ship B (answer is 315 degrees) c)the bearing of ship B from ship A (answer is 135 degrees) 翻譯: A同B兩隻船在同一時間離開一個口岸 船A以300度分向航行了12公里, 與此同時船B亦以150度方向航行了12公里. 求: a)兩船距離 b)由船B到船A的分位 c)由船A到船B的分位 已在上文註明了答案 希望有數學高人就以上答案做d steps 只徵求steps!! 一題數學題20fun(x2)抵到爛 識既快d答我 最快答到三題即送分!!
最佳解答:
Have you learned sine rule and cosine rule? If no, method to solve will be different. 2009-08-07 07:34:57 補充： (a) Assume you learned cosine rule. Let the port be P, so AP = BP = 12 km. Angle APB = 150 degree so AB^2 = AP^2 + BP^2 - 2(AP)(BP) cos (angle APB) = 12^2 + 12^2 - (2)(12)(12)cos 150. = 144 + 144 - 288 cos 150 so AB = sqrt ( 288 - 288 cos 150) = 23.18 = 23.2 km. (b) Since AP = BP, so triangle APB is an isos. triangle, so angle PAB = angle PBA = (180 - 150)/2 = 30/2 = 15 degree. so bearing of A from B = 360 - 15 - (90 - 60) = 360 - 15 - 30 = 360 - 45 = 315 degree. (c) Bearing of B from A = 90 + (90 - 60) + 15 = 90 + 30 + 15 = 135 degree.
其他解答:
”三角的應用”數學題一條 20fun x2
發問:
Two ships A and B leave a port at the same time.Ship A sails on a bearing of 300 degrees for 12km while ship B sails on a bearing of 150 degrees for 12km.Find:a)the distance between the two ships (answer is 23.2km)b)the bearing of ship A from ship B (answer is 315 degrees)c)the bearing of ship B from... 顯示更多 Two ships A and B leave a port at the same time. Ship A sails on a bearing of 300 degrees for 12km while ship B sails on a bearing of 150 degrees for 12km. Find: a)the distance between the two ships (answer is 23.2km) b)the bearing of ship A from ship B (answer is 315 degrees) c)the bearing of ship B from ship A (answer is 135 degrees) 翻譯: A同B兩隻船在同一時間離開一個口岸 船A以300度分向航行了12公里, 與此同時船B亦以150度方向航行了12公里. 求: a)兩船距離 b)由船B到船A的分位 c)由船A到船B的分位 已在上文註明了答案 希望有數學高人就以上答案做d steps 只徵求steps!! 一題數學題20fun(x2)抵到爛 識既快d答我 最快答到三題即送分!!
最佳解答:
Have you learned sine rule and cosine rule? If no, method to solve will be different. 2009-08-07 07:34:57 補充： (a) Assume you learned cosine rule. Let the port be P, so AP = BP = 12 km. Angle APB = 150 degree so AB^2 = AP^2 + BP^2 - 2(AP)(BP) cos (angle APB) = 12^2 + 12^2 - (2)(12)(12)cos 150. = 144 + 144 - 288 cos 150 so AB = sqrt ( 288 - 288 cos 150) = 23.18 = 23.2 km. (b) Since AP = BP, so triangle APB is an isos. triangle, so angle PAB = angle PBA = (180 - 150)/2 = 30/2 = 15 degree. so bearing of A from B = 360 - 15 - (90 - 60) = 360 - 15 - 30 = 360 - 45 = 315 degree. (c) Bearing of B from A = 90 + (90 - 60) + 15 = 90 + 30 + 15 = 135 degree.
其他解答:
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