# Word problem involving a ship?

A ship at A is to sail to C 56 kilometer north and 258 km east of A. After sailing North 25 degrees 10 minutes east for 120 miles to P the ships is headed toward C. Find the distance of P from C and the required course to reach C.

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I do not know how to start solving it. The find part is just how the problem is written in my text. – Latino Heat Jul 7 '12 at 1:57
Are you doing trigonometry of the plane, or are you doing spherical trigonometry? If you are doing trigonometry of the plane, you are probably expected to assume that the Earth is flat. If distances are not too large, this gives a reasonably good approximation. Because the data mention minutes, you may be expected to do a spherical trig solution. – André Nicolas Jul 7 '12 at 2:03
The question is hard to understand. Can you check to see that you have copied it out exactly is it was written, or give us the reference so maybe we can see for ourselves, or put an image up somewhere? – Gerry Myerson Jul 7 '12 at 4:37
To answer the question in your title: Yes, what you wrote is a word problem involving a ship. – celtschk Jul 7 '12 at 20:59

I'll assume that you're working in the plane, rather than on the sphere. If that's not the case, skip this answer. Let's put some coordinates on your picture. Using $(east, north)$ coordinates, assume the ship starts at $A=(0,0)$ and sails to $P = (p_e,p_n)$. Along with the north axis, you can draw a right triangle with the lower angle $\alpha$ equal to 25 degrees 10 minutes and hypotenuse $193.1218$ (converting miles to kilometers). Now you can use $193.1218\sin\alpha$ and $193.1218\cos\alpha$ to find the coordinates $p_e$ and $p_n$ (do you see how?). Now from $P$ draw two lines, one to $C$ and one in the east direction. You can make another right triangle and using the (now known) lengths of the east and north sides you can find the heading angle and the distance to $C$. Hope this hint helps.
We are given the starting and ending positions: $A(0,0)$ and $C(258,56)$. Suppose the angle $\theta=25^0 10^'$. Then the intermediate position of $P$ is $(120 \cos(\theta), 120 \sin(\theta))=(108.609,51.030)$ and so now one can determine the distance $PC$ and the angle (heading) from $P$ to $C$