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January 15th, 2016, 09:47 AM   #1
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Bearings? Help

If a habour H is 25km due North of an airport A. Town T is 50km due East of H. So what is the bearing of T from A?

I'm having trouble with working it out. Please help.

Last edited by skipjack; January 15th, 2016 at 02:00 PM.
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January 15th, 2016, 12:59 PM   #2
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Let a be the angle of interest. a=arctan(2). a is the angle counterclockwise from due north.
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January 15th, 2016, 01:08 PM   #3
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The first thing to do is make a sketch. I'm assuming you've already done that.

Now, note that $\triangle AHT$ is a right triangle with $\angle H$ as the right angle. We want to find $\theta = \angle HTA$.

We find $\tan \theta = \dfrac{|HA|}{|HT|} = \dfrac{25}{50} = \dfrac{1}{2}$

And thus $\theta = \tan^{-1}\left(\dfrac{1}{2}\right)\approx26.57^\circ$

Now you just have to write it as a bearing.
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January 15th, 2016, 01:18 PM   #4
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The answer is 63.4, but can I have the working?

Last edited by skipjack; January 15th, 2016 at 02:03 PM.
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January 15th, 2016, 01:29 PM   #5
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The bearing of T from A is $\displaystyle (\tan^{-1}(2))^\circ$ due east from north. As previously suggested, make a diagram.
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January 15th, 2016, 02:06 PM   #6
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The answer is 63.4 degrees to 3 significant figures (because tan of the bearing is 2).
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January 15th, 2016, 02:16 PM   #7
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Didn't someone say, "make a sketch" ? ... makes the whole problem rather elementary if you paid heed to their good advice.
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January 16th, 2016, 05:14 AM   #8
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So what would be the bearing from A to T?
I think it would be 243 but I don't know how to work it out?

Last edited by skipjack; January 16th, 2016 at 05:42 AM.
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January 16th, 2016, 05:40 AM   #9
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Stick with 63.4 degrees. There's no need to add on a further 180 unless you want the bearing of A from T, which isn't what you asked for originally.
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January 16th, 2016, 05:57 AM   #10
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I know but i was curious of what A to T was so how would I work that out?
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