Asked by Dragon

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.

Answers

Answered by MathMate
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<sup>-1</sup>(x/y) W
Answered by hipo
you hipocrit saying don't breack connexus rules? cmon havesome sence
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