document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1168157>Question 1168157</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Two ships leave a port O. <BR>\n" );
document.write( "One ship travels on a bearing of 340° to a point P which is 50 km from O. <BR>\n" );
document.write( "The other ship travels on a bearing of 060° to a point Q, 85 km from O. <BR>\n" );
document.write( "(a) draw a diagram to represent the position of the port and the two ships. On your diagram carefully label north, the given angles and the distance travelled.<BR>\n" );
document.write( "(b) calculate the distance PQ in km.<BR>\n" );
document.write( "(c) Determine the bearing of P from O. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #792803 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=792803\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Draw this \r<BR>\n" );
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document.write( "Law of cosines with a being OP and b being OQ,  c is PQ.<BR>\n" );
document.write( "c^2=a^2+b^2-2abcosC<BR>\n" );
document.write( "=50^2+85^2=2(4250)cos 80<BR>\n" );
document.write( "=2500+7225-1476<BR>\n" );
document.write( "=8248.99=8249<BR>\n" );
document.write( "c=sqrt(8249)<BR>\n" );
document.write( "=90.82 or 91 km.\r<BR>\n" );
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document.write( "law of sines for angle Q<BR>\n" );
document.write( "sin 80/90.82=sin x/50<BR>\n" );
document.write( "sin x=0.5421<BR>\n" );
document.write( "take arc sin of that<BR>\n" );
document.write( "x=32.83 deg \r<BR>\n" );
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document.write( "It can be shown on the drawing that the triangle in the second quadrant has angles 20, 67.17, and therefore 92.83 deg<BR>\n" );
document.write( "That means the part of the triangle between the horizontal line at the y-axis to the bearing has to be 2.83 degrees.<BR>\n" );
document.write( "The bearing from point Q (I think you meant that and not point O, which would simply bear 340, since it is given) is 270 deg + that 2.83 deg or 272.83 deg.\r<BR>\n" );
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document.write( "Alternatively, Point Q is 42.5 km N of point O and point P is 50 sin 70 or 46.98 km north of point O. PQ is the hypotenuse of a triangle with vertical distance 4.5 miles.<BR>\n" );
document.write( "sine of the angle is 4.5/90.82, and the angle can be shown to be 2.84 degrees. </SPAN>\n" );
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