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April 2nd, 2008, 09:36 AM   #1
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Sine Law: Ship Question

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Three ships, A, B, and C, are sailing along the same straight line on a course of 240°. Ship A is at the front, and ship C bring up the rear. From ship A, the bearing of a navigation buoy is 025°. The buoy is 2.5 km from ship A and 2km from ships B and C. To the nearest degree, what is the bearing of the buoy from ship C? from ship B?
Okay, so I've been doing this question for the past couple of days and I'm having no luck getting the answer... If someone could assit me with this I'd be so happy. I've attached an image (what I understand it to be, it's not necessarly correct)...

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millerst is offline  
 
April 2nd, 2008, 05:48 PM   #2
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[color=darkblue]Greetings:

Given a course of 240 deg, the ray extending from the origin makes a 60 deg angle with the negative x-axis (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
nikkor180 is offline  
April 2nd, 2008, 10:07 PM   #3
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This is not a physics problem, is it?
johnny is offline  
April 3rd, 2008, 08:36 AM   #4
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Quote:
Originally Posted by nikkor180
[color=darkblue]Greetings:

Given a course of 240 deg, the ray extending from the origin makes a 60 deg angle with the negative x-axis

Rich B.[/color]
rmath4u2@aol.com
I didn't check the rest of your work, but from what I remember from my trig bearing unit, bearing is measured clockwise from the positive y axis; what we would visualize as "North". The ray sweeps clockwise through Q1, then QIV, then QIII, and then QII. If this is true, then the ray would actually make a 30 degree angle with the negative x axis.

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)
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April 3rd, 2008, 09:39 AM   #5
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[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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