Friday

December 19, 2014

December 19, 2014

Posted by **Dragon** on Wednesday, January 26, 2011 at 8:24pm.

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.

- Math (Pre-cal) -
**MathMate**, Wednesday, January 26, 2011 at 8:59pmWhat 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

**Answer this Question**

**Related Questions**

math - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...

math - At noon, ship A is 30 nautical miles due west of ship B. Ship A is ...

CAL - At noon, ship A is 30 nautical miles due west of ship B. Ship A is sailing...

Cal 1 - (1 pt) At noon, ship A is 50 nautical miles due west of ship B. Ship A ...

math - At noon, ship A is 20 nautical miles due west of ship B. Ship A is ...

math - At noon, ship A is 50 nautical miles due west of ship B. Ship A is ...

Calculus - At noon, ship A is 10 nautical miles due west of ship B. Ship A is ...

Calc - At noon, ship A is 40 nautical miles due west of ship B. Ship A is ...

math - At noon, ship A is 40 nautical miles due west of ship B. Ship A is ...

math - At noon, ship A is 40 nautical miles due west of ship B. Ship A is ...