Originally Posted by CapnHand
Ok, so this is simpler than I first thought. This is way I should have done it the first time.
First of all, this is the situation. The distance the friend’s boat will travel is 5t (t = hours) NM on a bearing of 036 Deg T. You are 84NM from him at a bearing of 096 Deg T relative to his boat. In order to meet, you will travel a distance of 7t NM at some unknown heading.
It’s obvious that the angle between the track of your friend’s boat and the bearing from his boat to yours is 96 – 36 = 60 Deg
Remember the law of sines? It’s handy. It says that for any triange the ratios of the length of each side to the angle opposite that side are equal.
If A is the angle between the heading you need and the bearing from your boat to your friends,
that means: sin60 / 7t = sinA / 5t
or sin –1 (5/7 * sin 60) = A
A = 38.2 Deg
Your heading then should be 96 + 38.2 + 180 = 314.2 Deg T
How long to get there?
We all remember that the sum of all the internal angles of a triangle = 180 so the remaining angle is 180 – 38.2 – 60 = 81.8 Deg
From the law of sines, 7t = 84 * sin 60 / sin 81.8
t =10.5 hours
While I applaud your attempt to answer the question, I see it as seriously flawed.
Given that your distance from his boat is 84NM to start with, and his is moving away from you, and you only move at 7 knots... I don't see how the answer could be 10.5 hours.
Draw the vectors... In 10.5 hours you have gone only 73.5nm on a heading of 73.8˚T. His position has now become 52.5nm @ 36˚T + the original 84NM @ 96˚T in the same 10.5 hours—119NM from your original position, or 45.5NM further along the same heading.
If you assume you were at coordinate (0,0) and have moved to (70,20.5)... boat B's position has gone from (83, -8.5) to (114, 33). You're still nowhere near them... Your solution assumes an interior angle of 60 degrees, but it has an interior angle of 120 degrees (96 -> 36)
. However, you are on a direct vector to their current position.
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