document.write( "Question 194865: Two ships leave harbor at the same time. The first sails North 15 degress West at 25 knots. The second sails North 32 degress East at 20 knots. After 2 hours, how far apart are the ships? \r
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document.write( "Is there a formula I have to follow? Thank you in advance. \n" );
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Algebra.Com's Answer #146210 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Two ships leave harbor at the same time. \n" ); document.write( "The first sails North 15 degress West at 25 knots. \n" ); document.write( "The second sails North 32 degress East at 20 knots. \n" ); document.write( "After 2 hours, how far apart are the ships? \n" ); document.write( ": \n" ); document.write( "From the given headings we can say the angle between the courses of the two ships: \n" ); document.write( "15 + 32 = 47 degrees \n" ); document.write( ": \n" ); document.write( "After two hrs \n" ); document.write( "the 1st ship will have traveled 2*25 = 50 mi \n" ); document.write( "the 2nd ship will have traveled 2*20 = 40 mi \n" ); document.write( ": \n" ); document.write( "We can use the law of cosines for a side/angle/side problem \n" ); document.write( "a^2 = b^2 + c^2 - (2bc * cos(A)) \n" ); document.write( ": \n" ); document.write( "Assign the values as follows \n" ); document.write( "a = distance between ships after 2 hr (side opposite angle A) \n" ); document.write( "b = 50 \n" ); document.write( "c = 40 \n" ); document.write( "A = 47 degrees \n" ); document.write( ": \n" ); document.write( "a^2 = 50^2 + 40^2 - (2*50*40 * cos(47)) \n" ); document.write( "a^2 = 2500 + 1600 - (4000*.682) \n" ); document.write( "a^2 = 4100 - 2728 \n" ); document.write( "a^2 = 1372 \n" ); document.write( "a = \n" ); document.write( "a = 37 mi apart \n" ); document.write( " |