document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1140522>Question 1140522</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi can you please help me with this problem. A ship leaves a port and travels with a bearing of 55 degrees for 100 km. It then travels 500 km on a course of 120 degrees. How far is the ship from the port? In order to return to the port, what is the bearing the ship must take? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #761056 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=761056\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> in the diagram below, the following occurs.\r<BR>\n" );
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document.write( "GH is the north south line for point A.\r<BR>\n" );
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document.write( "PM is the east west line for point A.\r<BR>\n" );
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document.write( "EF is the north south line for point B.\r<BR>\n" );
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document.write( "NO is the east west line for point B.\r<BR>\n" );
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document.write( "IK is the north south line for point C.\r<BR>\n" );
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document.write( "QL is the east west line for point C.\r<BR>\n" );
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document.write( "the bearing form point A to point B is 55 degrees which is the angle GAB.\r<BR>\n" );
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document.write( "the bearing from point B to point C is 120 degrees which is the angle EBC.\r<BR>\n" );
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document.write( "angle FBC is equal to 60 degrees because it is supplementary to angle EBC which is 120 degrees.\r<BR>\n" );
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document.write( "angle ABD is 55 degrees because it is complementary to angle BAD which is 35 degrees because it is complementary to angle GAB which is 55 degrees.\r<BR>\n" );
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document.write( "angle ABC is equal to 115 degrees because it is the sum of angles ABD and FBC which makes it the sum of 55 and 60 degrees.\r<BR>\n" );
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document.write( "side AB of triangle ABC is equal to 100 km because it is the distance from point A to point B.\r<BR>\n" );
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document.write( "side BC of triangle ABC is equl to 500 km because it is the distance from point B to point C.\r<BR>\n" );
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document.write( "you can use the law of cosines to find the length of side AC of triangle ABC.\r<BR>\n" );
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document.write( "the law of cosines says that b^2 = a^2 + c^2 - 2 * a * c * cos(B).\r<BR>\n" );
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document.write( "the law of cosines assumes that side a is opposite angle A and side b is opposite angle B and side c is opposite angle C.\r<BR>\n" );
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document.write( "to fit this formula, we let:\r<BR>\n" );
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document.write( "angle (B) = angle ABC = 115 degrees.<BR>\n" );
document.write( "side b = AC = what we want to find.<BR>\n" );
document.write( "side c = AB = 100 kilometers.<BR>\n" );
document.write( "side a = BC = 500 kilometers.\r<BR>\n" );
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document.write( "b^2 = a^2 + c^2 - 2 * a * c * cos(B) becomes:\r<BR>\n" );
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document.write( "b^2 = 500^2 + 100^2 - 2 * 500 * 100 * cos(115).\r<BR>\n" );
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document.write( "solve for b^2 to get b^2 = 302261.8262.\r<BR>\n" );
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document.write( "solve for b to get b = 549.7834357.\r<BR>\n" );
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document.write( "that's the direct distance from point C back to the port at point A.\r<BR>\n" );
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document.write( "we can now use the law of sines to find angle C.\r<BR>\n" );
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document.write( "law of sines says that b / sin(B) = c / sin(C).\r<BR>\n" );
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document.write( "we know:\r<BR>\n" );
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document.write( "b = 549.7834357<BR>\n" );
document.write( "c = 100<BR>\n" );
document.write( "angle B = 115 degrees.\r<BR>\n" );
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document.write( "b / sin(B) = c / sin(C) becomes:\r<BR>\n" );
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document.write( "549.7834357 / sin(115) = 100 / sin(C)\r<BR>\n" );
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document.write( "solve to get sin(C) = c * sin(B) / b = 100 * sin(115)  549.7834357 = .1648481435.\r<BR>\n" );
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document.write( "solve to get angle C = arcsin(.1648481435) = 9.488412932 degrees.\r<BR>\n" );
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document.write( "since triangle  BFC is a right triangle and angle FBC is equal to 60 degrees, then angle BCF is equal to 30 degrees because angle BCF is complementary to angle FBC.\r<BR>\n" );
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document.write( "angle ACF is equal to 30 degrees minus 9.488412932 degrees = 20.51158707 degrees.\r<BR>\n" );
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document.write( "angle ICA (shown as angle 4 in the diagram) is equal to 270 plus 20.51158707 = 290.51158707 degrees.\r<BR>\n" );
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document.write( "that's the bearing from point C to point A.\r<BR>\n" );
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document.write( "your solutions are (if i did this correctly):\r<BR>\n" );
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document.write( "the direct distance from point C back to the port at point A is 549.7834357 kilometers.\r<BR>\n" );
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document.write( "the bearing from point C back to the port at point A is 290.51158707 degrees.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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