September 11th, 2012, 08:00 AM  #1 
Senior Member Joined: Sep 2012 Posts: 201 Thanks: 1  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. Answer = 025 degrees. 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 
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 northsouth lines carefully... you should see that 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 9054 is that the correct way of doing the calculation?

September 11th, 2012, 12:00 PM  #4  
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Bearings Quote:
"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 . Calling the distance from B to S, "y", we have . (Equivalently, and . (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 . 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  
Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233  Re: Bearings Quote:
Yes, from A clockwise from north the angle is 54º so angle SAB = 9054 = 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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