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Question: A cargo ship leaves port at 8:30 am going south at 20 knots. At the same time, a speed boat leaves port going east at 40 knots. Find the time of day, to the nearest minute,when the distance between the two boats will be approx. 536.66 nautical miles?

I'm not sure how to answer this question. This is not a trig question so I wouldn't be able to solve it that way. I created a triangle to show the traveling but what confusing me is the time. I'm thinking since it's a distance question I would need the distance formula but I'm not entirely sure how I would set that up. How would I solve this one? I maybe over thinking it.

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You have a right triangle. One leg is $20t$, the other is $40t$. You want the hypotenuse to be $536.66$. The Pythagorean theorem gives you a quadratic in $t$

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    Oh so like me saying $(20t)^2 + (40t)^2= (536.66)^2$? I got 12 for $t$ so does saying the answer is 8:30pm sound right? – 2017-02-12
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    Yes, that is right – 2017-02-12
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    Great! Yeah I was really overthinking this question. Thank you for you help :). – 2017-02-12
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    When the distance traveled is only off by a factor of 6.41 from the radius of the earth, I think it becomes relevant to consider the fact that earth is spherical, and therefore the Pythagorean theorem is not correct since the space is not flat. At smaller scales it could be ignored, but 536.66 nmi is not small. – 2017-02-12
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    @Rodolvertice If you do consider curvature what answer do you get? My guess is that to the nearest minute (as mentioned in the question) it does not matter. – 2017-02-12
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    @Rodolvertice: a good point. Do you think we want the great circle distance or the Euclidean (straight line through the earth) distance? I am sure the problem intends you to ignore the curvature. Using $21600\ nm$ as the circumference of the earth, the $480 nm$ represents $0.1396$ radian. The curvature brings the end points $2.45$ nm closer together, so if we use Euclidean distance it is several minutes of error – 2017-02-12
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Because each boat is going at a constant speed in a constant direction, you can solve this very easily by proportions. Find out how far apart the boats are in one hour (this is the hypotenuse of a right triangle). Divide $536.66$ by that distance; that tells you the factor how much you have to scale up your one-hour triangle to find the final positions of the boats.

The factor by which you have to scale up the distance is also the factor by which you have to scale up the time, that is, it's the number of hours that have passed since the boats started.

Ross Millikan's answer is more general, however, because the idea of it also applies when the boats' speeds are known but not constant. (The basic idea is, how far has each boat gone at time $t$ hours since they started?)

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    That's very true. This is a very good approach as well. – 2017-02-12