bearing (1) quick

2009-11-14 2:00 am
1. A ship sails from point A at a bearing of 155° to poinr B. If it has to sail from point B back to point A, what direction should the ship sail? (ANS. 335° /N25°W)

2.Two ships B & C depart from port A at ther same time. Ship B sails 8 km at a bearing of S45°W and then then stops. Ship C sails 12 km at a bearing of S57°E and then stops. After both stopped,
(A)what is the distance between ships B and C? (15.7km)
(B)find the compass bearing of ship B from ship C. (N86.8°W)

3.Jacky walks 2 km due east and then 3 km due south. Find the true bearing of his final position from the starting point.

4.A man starts at O and walks 2 km at a bearing of N27°E to A, and then turns 90° to his right and walks 1 km to B. What is the compass bearing of B from O?

5.There are two routes to go from A to B. The first route is to go 350m at a bearing of N50°W and then 1000m at a bearing of N40°E. The second route is to go east and then north.
(a) Find the length of the second route.
(b) Find the difference in length between these two routes.

回答 (3)

2009-11-14 8:01 am
✔ 最佳答案
(1) The bearing of B from A is 155 deg or S25deg E
The bearing of A from B is N25degW or 335deg
(2) (A) The East-West distance between B and C is 8sin45 + 12sin57 = 15.72 km
The North-South distance between B and C is 12cos57 - 8cos45 = 0.8788 km
BC = √(15.72^2 + 0.8788^2) = 15.7 km
(B) Bearing of B from C is NxW where tan x = 15.72/0.8788 = 17.89
x = 86.8 degree
The bearing is N86.8W
(3) The compass bearing is SxE where tan x = 2/3
x = 33.7
Compass bearing = S33.7E
True bearing = 180 - 33.7 = 146.3 degrees
(4) OAB is a right angled triangle
tan AOB = 1/2
angle AOB = 26.6 degrees
The compass bearing of B from O is N27 + 26.6E or N53.6E
(5) (a) The east distance = -350sin50 + 1000sin40 = 374.7 m
The north distance = 350cos50 + 1000cos40 = 991.0m
Length of second route = 374.7 + 991.0 = 1366m
(b) Difference in length = 1366 - 350 - 1000 = 16m
2009-11-14 3:24 am
no any pics in my book
2009-11-14 3:01 am
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