SOLUTION: Please help. Thank you.
A ship leaves port and travels for 50 km along a bearing of N 25 degree W, then turns along a bearing of S 70 degree W for 100 km, followed by a 75 km leg
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-> SOLUTION: Please help. Thank you.
A ship leaves port and travels for 50 km along a bearing of N 25 degree W, then turns along a bearing of S 70 degree W for 100 km, followed by a 75 km leg
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Question 1113500: Please help. Thank you.
A ship leaves port and travels for 50 km along a bearing of N 25 degree W, then turns along a bearing of S 70 degree W for 100 km, followed by a 75 km leg along a bearing of S 20 degree E, and finally along a bearing of N 80 degree E for 80 km.
a) where is ship's position relative to the port?
b) How far is the ship from the port?
c) Along what bearing must the ship travel to return to port? Answer by Theo(13342) (Show Source):
add up all the north direction distances and subtract all the south direction distances to get a net distance that is either north or south.
add up all the east direction distances and subtract all the west direction distances to get a net distance that is either east or west.
the ship starts at point A.
that would be the origin on the graph.
it travels 50 kilometers to point B on a bearing of north 25 degrees west.
this forms right triangle BAH where the north distance is 50 * cos(25) and the west distance is 50 * sin(25).
the north distance is equal to AH which is equal to 50 * cos(25) which is equal to 45.31538935 kilometers.
the west distance is equal to HB which is equal to 50 * sin(25) which is equal to 21.13091309 kilometers.
what that looks like is in the following graph.
from point B, 100 kilometers on a bearing of south 70 degrees west.
this forms right triangle BCI where the south distance is 100 * cos(70) and the west distance is 100 * sin(70).
the south distance is equal to BI which is equal to 100 * cos(70) which is equal to 34.20201433 kilometers.
the west distance is equal to IC which is equal to 100 * sin(70) which is equal to 93.96926208 kilometers.
what that looks like is in the following graph:
from point C, he travels 75 kilometers on a bearing of south 20 degrees east.
this forms right triangle CDJ where the south distance is 75 * cos(20) and the east distance is 75 * sin(30).
the south distance is equal to CJ which is equal to 75 * cos(20) which is equal to 70.47694656 kilometers.
the east distance is equal to JD which is equal to 75 * cos(20) which is equal to 25.65151075 kilometers.
what that looks like is in the following graph:
from point D, he travels 80 kilometers to point F on a bearing of north 80 degrees east.
this forms right triangle DFK where the north distance is 80 * cos(80) and the east distance is 80 * sin(80).
the north distance is equal to DK which is equal to 80 * cos(80) which is equal to 13.89185421 kilometers.
the east distance is equal to 80 * sin(80) which is equal to KF which is equal to 78.78462024 kilometers.
what this looks like is in the following graph:
if we consider the distance going north to be positive and the distance going south to be negative, we get a net distance of 50 * cos(25) km - 100 * cos(70) km - 75 * cos(20) km + 80 * cos(80) which is equal to -45.47171733.
this means he has traveled a net distance of 45.47171733 kilometers directly south of his home port.
if we consider the distance going west to be negative and the distance going east to be positive, we get a net distance of -50 * sin(25) - 100 * sin(70) + 75 * sin(20) + 80 * sin(80) which is equal to -10.66404418.
this means he has traveled a net distance of 10.66404418 kilometers directly west of his home port.
this forms right triangle FLA which is our final triangle.
the distance from point F to L is equal to FL which is equal to 45.47171733 kilometers.
the distance from point L to point A is equal to LA which is equal to 10.66404418 kilometers.
the distance from point A to point F is equal to FA which is the hypotenuse of right triangle FAL and is therefore equal to square root of ((45.47171733)^2 + (10.66404418)^2) which is equal to 46.70544845 kilometers.
the net bearing from point F to point A would be arc tangent (LA / FL) which is equal to arc tangent (10.66404418 / 45.47171733) which is equal to north 13.19850608 degrees east.
what this looks like is shown in the following graph:
the answers to your questions are, as best i can determine, are shown below:
a) where is ship's position relative to the port?
the ship is 45.47171733 kilometers south of the home port and 10.66404418 kilometers west of the home port.
b) How far is the ship from the port?
the ship is 46.70544845 kilometers from the home port as measured by the straight line distance between the final destination of the ship and the home port.
c) Along what bearing must the ship travel to return to port?
the ship would have to travel on a bearing of north 13.19850608 degrees east to get back to the home port from its final destination.
the overall graph of the ship's route is shown in the following graph:
