April 2nd, 2008, 09:36 AM  #1  
Newbie Joined: Apr 2008 Posts: 1 Thanks: 0  Sine Law: Ship Question Quote:
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April 2nd, 2008, 05:48 PM  #2 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Greetings: Given a course of 240 deg, the ray extending from the origin makes a 60 deg angle with the negative xaxis (note that such information implies the diagram is not to scale). Thus angle(BAO) has measure 5 deg, where B, A and O denote the locations of Buoy, shop A, and the origin respectively. Letting C represent the location of ship C, the law of sines gives us, sin(5 deg) / 2 = sin(angle BCA) / 2.5 ==> sin(BCA) = 1.25*sin(5 deg) ==> measure of BCA = sin^1(1.25*sin(5)) ~~ 6.25 deg. I shall assume you can take it from there. Note: You may wish to use the fact that base angles of an isosceles triangle are congruent. Regards, Rich B.[/color] rmath4u2@aol.com 
April 2nd, 2008, 10:07 PM  #3 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0 
This is not a physics problem, is it?

April 3rd, 2008, 08:36 AM  #4  
Member Joined: Feb 2008 Posts: 44 Thanks: 0  Quote:
Q I = 90 degrees Q IV = 90 degrees 240  90  90 = 60 degrees (the remainder of the angle lies in QIII) 90  60 = 30 degrees (the remaining angle from above, subtracted from the number of degrees per quadrant)  
April 3rd, 2008, 09:39 AM  #5 
Member Joined: Feb 2008 Posts: 89 Thanks: 0 
[color=darkblue]Julien: That I did not know. So, I guess the diagram is way off! Regards, Rich B.[/color] rmath4u2@aol.com 

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