SOLUTION: The bearings of ships A and B from a port P are 225° and 116° respectively. Ship A is 3.9 km from ship B on a bearing of 258º. Calculate the distance of ship A from P.

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Question 1177481: The bearings of ships A and B from a port
P are 225° and 116° respectively. Ship A is
3.9 km from ship B on a bearing of 258º.
Calculate the distance of ship A from P.
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website!
the following pictures should tell the story.

picture 1 shows the bearing from P to B and from P to A and from B to A.

that's 116 degrees and 225 degrees and 258 degrees respectively.

picture 2 shows the additional angles that results from the angles in picture 1.

angle FPB = 26 degrees because 116 degrees is equal to 90 degrees plus 26 degrees.

angle APB = 109 degrees because it's the difference between 225 degrees and 116 degrees.

angle PBG - 26 degrees because it is the alternate interior angle of angle FPB of the transversal PB between parallel lines PF and GB.

angle ABG = 12 degrees because it is is the diffence between a 258 degree angle and a 270 degree angle.

picture 3 shows the triangle formed with all its interior angles.

because angle APB = 109 degrees and because angle PBA = 38 degrees, angle PAB has to be 33 degrees because the sum of the interior angles of a triangle = 180 degrees and 180 minus 109 minus 38 = 33 degrees.

picture 3 also shows the length of AB to be equal to 3.9 kilometers.

from picture 3, you can use the law of sines to get:

3.9 / sin(109) = PA / sin(38)

solve for PA to get:

PA = 3.9 * sin(38) / sin(109) = 2.539431605 km.

that should be your answer if i did this correctly.

the hard part was getting the angles right.
once that was gone, it was a simple application of the law of sines.

3.9 km was the length of the side opposite angle 109.
PA was the side opposite angle 38.
we needed to find the length of side PA.