What program are you looking at?
The calculations can be done by hand.
The ships sails due west for two hours at 20 knots, so by 14:00, it is 40 knots west of port.
It then changes course to N54°W for one hour. So it has travelled further west 20*tan(54°) and towards north 20cot(54°).
So total distance due west
x = -(40+20tan(54°))
and due north
y = 20cot(54°)
Thus bearing at 15:00 is
N tan-1(x/y) W
I can't seem to figure this out.
A ship leaves at noon and travels due West at 20 knots. At 2:00 it changes to N54°W...Find the bearing and the distance FROM port at 3:00 pm.
I've looked at this program for a while but can't figure it out.
2 answers
you hipocrit saying don't breack connexus rules? cmon havesome sence