document.write( "Question 1110434: Two ships leave Liverpool at the same time. One of them travels north-west at an average speed of 10.5 km/h while the other travels at an average speed of 14 km/h on a bearing of 280ᐤ. How far apart are these ships after 2 hours?\r
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document.write( "I do not understand this because, unless I'm mistaken, a bearing angle of 280ᐤ is going north-west, and the question doesn't say what angle the first ship is going on. I also do not know how to figure out how far apart they are after two hours. Help would be much appreciated! \n" );
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Algebra.Com's Answer #725439 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! Two ships leave Liverpool at the same time. One of them travels north-west at an average speed of 10.5 km/h while the other travels at an average speed of 14 km/h on a bearing of 280ᐤ. How far apart are these ships after 2 hours? \n" ); document.write( ": \n" ); document.write( "Exactly northwest is 315 degrees so the angle between the two ship's path is: \n" ); document.write( "315-280 = 35 degrees \n" ); document.write( "Find the distance each ship has traveled in two hrs; dist = time * speed \n" ); document.write( "2*10.5 = 21 km \n" ); document.write( "and \n" ); document.write( "2*14 = 28 km \n" ); document.write( "We have a triangle we can solve; the enclosed angle is 35 degrees (A), has legs of 21 and 28 km. the distance between the ships is the side opposite this angle a \n" ); document.write( "Use the law of cosines a^2 = b^2 + c^2 - 2(b*c)cos(A) \n" ); document.write( "In this problem \n" ); document.write( "a^2 = 21^2 + 28^2 - 2(21*28)Cos(35) \n" ); document.write( "a^2 = 441 + 784 - 2(588)*.819 \n" ); document.write( "do the math \n" ); document.write( "a = \n" ); document.write( "a = 16.17 km apart after two hours \n" ); document.write( ": \n" ); document.write( "this is the method, I'll let you check my math. \n" ); document.write( " |