”三角的應用”數學題一條 20fun x2

用戶221926

2009-08-07 9:48 am

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!!

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回答 (1)

用戶10012

2009-08-07 3:34 pm

✔ 最佳答案

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.

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