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Two ships have radio equipment with a range of 200 km.One is 155km North42 degrees 40 minutes east and the other is 165 km north 45 degrees 10 minutes west of a shore station. Can the two ships communicate directly?

How would I solve this problem I know I have to make a triangle but I am not sure how the triangle would look like.

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Where is the share station? (OK, it's at least 165 km away from the North pole :-)) – celtschk Aug 13 '12 at 17:36
ooops shore station is what I meant. – Fernando Martinez Aug 13 '12 at 17:39
What is "155km North42 degrees 40 minutes east"? Is it a range of 155km and some angle? If so, what is the angle measured from (from North toward East, for example)? – Ross Millikan Aug 13 '12 at 17:41
That still doesn't tell where this station sits. The distance of one degree west differs a lot on whether you are near the equator or near the pole. – celtschk Aug 13 '12 at 17:42
@RossMillikan: Usually if "degree" is combined with "east" or "west", it means longitude. That is, the angle is measured from the pole. – celtschk Aug 13 '12 at 17:44
up vote 1 down vote accepted

We will assume that the Earth is flat, and that we are not too far North. Draw a picture. Let the share station be at $C$. For the first ship, go $42^\circ40'$ East from due North, and travel $155$ km. Call the resulting point $A$. So we face North from $C$, turn $42^\circ40'$ clockwise, and sail $155$ km.

For the second, go $45^\circ10'$ West from due North, and travel $165$ km. Call the resulting point $B$.

Then $\triangle ABC$ has $CA=155$, $CB=165$, and $\angle C=42^\circ40'+45^\circ10'=87^\circ 50'$.

By the Cosine Law, $$(AB)^2=155^2+165^2-2(155)(165)\cos C.$$ Calculate. It turns out that the distance is about $222$ km.

Remark: We can proceed more informally without the Cosine Law, and with a little crossing of the fingers. Note that $\angle C$ is almost a right angle. If we pretend it is a right angle, we can use the Pythagorean Theorem to estimate the distance. That gives $226$ km. Not very different from $222$.

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yes your work must be correct for its the same answer in my book, thanks. – Fernando Martinez Aug 13 '12 at 17:52
In principle, my calculation and the one in the book are both in principle not right. If we were talking about sailing $2550$ and $2650$ km, the curvature of the Earth would make a significant difference, and we would have to use spherical trigonometry. – André Nicolas Aug 13 '12 at 17:56
I see I have not heard of this spherical trigonometry because I think the focus of the text I am using is plane trigonometry, so your calculation is probably the one the book I am using intended me to do. – Fernando Martinez Aug 13 '12 at 17:58
If you have covered the Cosine Law, you are probably expected to use that. If Cosine Law has not been covered, you are probably just supposed to notice that $C$ is almost a right angle. – André Nicolas Aug 13 '12 at 18:05
yes cosine law is the focus of the section where I got that problem from and it probably was what it wanted me to do. The problem for me is when they give out the information in sentences its easier to solve if they just say a=this A=this angle. – Fernando Martinez Aug 13 '12 at 18:10

With the shore station as the common vertex there should be two right angled triangles, one with angle 42 degrees 40 minutes and other with angle 45 degrees 10 minutes to yhe otjer side since it is west.

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