September 11th, 2012, 09:00 AM  #1 
Senior Member Joined: Sep 2012 Posts: 200 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, 09:56 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, 11:38 AM  #3 
Senior Member Joined: Sep 2012 Posts: 200 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, 01:00 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, 01:47 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 12th, 2012, 12:49 AM  #6 
Senior Member Joined: Sep 2012 Posts: 200 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. 

Tags 
bearings 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Trigonometry and bearings  sakuraa  Trigonometry  6  February 25th, 2013 09:49 AM 
degrees and bearings, am i right?  mathslog  Algebra  2  April 29th, 2012 12:25 PM 
Trig. Bearings  Sunyata.  Trigonometry  1  October 29th, 2011 02:28 AM 
Help on bearings  a.sundar23  Algebra  1  May 5th, 2009 07:46 PM 
Pytragoras and Bearings need help  SuperMaths  Algebra  1  January 12th, 2009 02:02 PM 