SOLUTION: A ship is 30 nautical miles from port, the bearing from the ship to port is N20°E.
a) What is the bearing from port to the ship?
b) If the ship sails 10 nautical miles directly
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-> SOLUTION: A ship is 30 nautical miles from port, the bearing from the ship to port is N20°E.
a) What is the bearing from port to the ship?
b) If the ship sails 10 nautical miles directly
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Question 1140759: A ship is 30 nautical miles from port, the bearing from the ship to port is N20°E.
a) What is the bearing from port to the ship?
b) If the ship sails 10 nautical miles directly north, how close to the port will it be? What is the new bearing from the ship to the port? Answer by Theo(13342) (Show Source):
bearing from the ship to the port is angle DAB = 20 degrees.
DBAC forms a rectangle.
all interior angles of a rectangle are each equal to 90 degrees.
angle ABC is equal to 20 degrees because alternate interior angles of a rectangle are equal (a rectangle is a parallalogram).
angle DBA = 70 degrees because 90 - 20 = 70.
the bearing from point B (the port) to point A (the ship) is equal to 70 degrees south west.
the north south lines are FG and EH.
the east west lines so far are DB and AC.
answer to question A is 70 degrees south west.
point A is the ship.
point B is the port.
because this is a rectangle, the length of DB is the same as the length of AC and the length of DA is the same as the length of BC.
check my next diagram and then read the notes below.
this is the same as the first diagram with the addition of line segment IJ which is the east west line 10 miles north of point A and parallel to line segment KL.
in this diagram, you have:
north south lines FG, BH.
east west lines NO, IJ, KL.
point M is 10 miles above point A which was the original location of the ship.
point M is the new location of the ship.
MB is the new distance from the ship to the port.
length of AD is the same length as BC which is equal to 28.19077862 miles.
length of MD is equal to the length of AD minus 10 = 18.19077862 miles.
that's the length of number 4 in the diagram.
point M and point D and point B form triangle MDB.
we know the length of DB and the length of DM.
we use pythagorus to find the length of MB which is equal to sqrt(DB^2 + MD^2) which is equal to sqrt(18.19077862^2 + 10.2606043^2) = 20.88502879 miles.
that's the distance from the ship to the port after the ship has moved north 10 miles.
the new bearing from the ship to the port is arctan(DB / DM) = arctan(10.2606043 / 18.19077862) = 29.42540014 degrees.
the new bearing from the ship to the port is north 29.42540014 degrees east.
the new distance from the ship to the port is number 5 on the diagram.
the new bearing from the ship to the port if number 6 on the diagram.
the numbers i am referring to are the numbers inside the circles on the diagram.
some additional information you might find useful.
bearing is the angle from the north south line rotating in a popsitive direction to the direction you are traveling.
navigation bearing is the north east and south west conventions used in this problem.
for example:
north east 20 degrees is a bearing of 20 degrees (diagram number 11).
south west 70 degrees is a bearing of 250 degrees (diagram number 12).
south east 70 degrees is a bearing of 110 degrees (diagram number 13).
north west 20 degrees is a bearing of 340 degrees (diagram number 14)..
the following diagram shows this relationship (diagram number 15).
