My Math Forum Bearings

 Algebra Pre-Algebra and Basic Algebra Math Forum

September 11th, 2012, 08:00 AM   #1
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Joined: Sep 2012

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Bearings

I have been stuck on a question for sometime, I am not 100% sure if its the books mistake or an error on my behalf. Here is the questions with the aid of diagrams of the questions.

Question: A & B are two points on a coastline. B is directly East of A. A ship S can be see from both A & B. The bearing from S to A is "054" degrees. The bearing of S from B is "324" degrees.

a) find the number of degrees in
i) angle SAB
ii) angle SBA

The answers for these two are i) 36 degrees and ii) 54 degrees. I cant really see how this works due to the fact that if your measure from S to A = 054 degrees it lies on a traversal which would add up to 180 degrees.

b) the stright line AB is 6KM. Workout the distance of the ship
i) from A
ii) from B
Answers: i) 6.6 and ii) 4.1 km

I have a feeling I am suppsoed to use SOHCAHTOA, but not 100% sure where to start.

d) Later the ship is seen 2.5KM from A with an angle ASB Still 90 degree. What is the bearing of the ship from A now.

This I have no idea where to start. If possible could some give me some guidence. I keep looking in the study books and internet and seem to confuse myself more.

c) Later the ship is seen 2.5km
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 September 11th, 2012, 08:56 AM #2 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233 Re: Bearings I think your diagram can be made to work if you do the following... a) i) At S start from the south line and measure 54° clockwise ii) At B start from the north line and measure 324° clockwise b) Yes, use SOHCAHTOA, SB = 6sin36 and SA = 6cos36 the answers given don't agree here, in particular, 6.6 is impossible since the hypotenuse AB is 6 and angle ASB is 90° d) Draw a new diagram and label sides and north-south lines carefully... you should see that $cos^{-1}(2.5/6)= 65.38 = \ angle \ SAB$ so the bearing of S to A is 25° measured clockwise from the south line at S.
 September 11th, 2012, 10:38 AM #3 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Re: Bearings Can i ask how you cam up with, to measure from the south line. I always thought you go from the north clockwise? Also, when I did what you said and workout out the angle North on the A is 54 degrees, the only way I could get 36 degree is take 90-54 is that the correct way of doing the calculation?
September 11th, 2012, 12:00 PM   #4
Math Team

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Re: Bearings

Quote:
 A & B are two points on a coastline. B is directly East of A. A ship S can be see from both A & B. The bearing from S to A is "054" degrees. The bearing of S from B is "324" degrees.
I think you have miscopied this. The way this is described, the "bearing from S to A" and the "bearing from S to B" cannot be what are given. That, together with the fact that "S can be seen from both A and B", rather than that A and B can be seen from S, makes me believe that "the bearing from A to S is 54 degrees and the bearing from B to S is 323 degrees".

"Bearings", as given here, are always measured clockwise from due north. So the angle at A to S, measured from the "due North" is 54 degrees. If you draw a line "due East", from A to B, you get a triangle which has an interior angle, at A, of 90- 54= 36 degrees. Similarly, "due West" on such a scale is 3(90)= 270 degrees so the iterior angle of the triangle, at B,is 323- 270= 53 degrees. Of course, the third angle of that triangle, at S, is 180- 36- 54= 180- 90= 90 degrees. Yes, this does happen to be a right triangle! You are told that the distance between A and B, the hypotenuse of the right triangle, is 6 km. Calling the distance from A to S, "x", we have $sin(36)= \frac{x}{6}$. Calling the distance from B to S, "y", we have $sin(53)= \frac{y}{6}$. (Equivalently, $cos(54)= \frac{x}{6}$ and $cos(36)= \frac{y}{6}$.

(If that third angle had NOT been exactly 90 degrees, you could have used the "sine law". Calling the length of the side opposite angle A "a", the length of the side opposite angle B, "b", and the length of the side opposite angle S, "s", the sine law says $\frac{a}{sin(A)}= \frac{b}{sin(B)}= \frac{s}{sin(S)}$. Given the three angles and one side, you can use that to solve for the other two sides. But since this happens to work out nicely to a right triangle, I suspect you have not yet had that.)

September 11th, 2012, 12:47 PM   #5
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From: North America, 42nd parallel

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Re: Bearings

Quote:
 Originally Posted by taylor_1989_2012 Can i ask how you cam up with, to measure from the south line. I always thought you go from the north clockwise? Also, when I did what you said and workout out the angle North on the A is 54 degrees, the only way I could get 36 degree is take 90-54 is that the correct way of doing the calculation?
I was trying to get the given numbers to work, so i noticed i could do that if i considered the appropriate half of the north-south line. Maybe they meant to write, 'the bearing from A to S' then i wouldn't need to use the south line at all.

Yes, from A clockwise from north the angle is 54º so angle SAB = 90-54 = 36

I am troubled by SA = 6.6, are you not troubled by it? Surely a side of a right triangle can't be longer than the hypotenuse...

 September 11th, 2012, 11:49 PM #6 Senior Member   Joined: Sep 2012 Posts: 201 Thanks: 1 Re: Bearings First I want to say thanks to both of you for the help. I did go over the question quite a few times last night, and came to the conclusion the only way I would get the 54 degrees is to draw a full N,E,S,W quadrants and work from there, which showed that the line intersects in the south, west quadrant. Which would give the correct answer. I did follow the trig and I understand. I personally think it is a miss written, I check the writing in the book, with which I wrote on the forum and did not miss type. Once again thanks for the help.

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