document.write( "Question 1136159: Two ships leave a harbor entrance at the same time. The first ship is traveling at a constant 10 miles per hour, while the second is traveling at a constant 14 miles per hour. If the angle between their courses is 123°, how far apart are they after 30 minutes? (Round your answer to the nearest whole number.)
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document.write( " = mi \n" );
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Algebra.Com's Answer #753913 by VFBundy(438)![]() ![]() You can put this solution on YOUR website! \r\n" ); document.write( "* *\r\n" ); document.write( " * *\r\n" ); document.write( " * * C\r\n" ); document.write( " * *\r\n" ); document.write( " 5 * *\r\n" ); document.write( " * * \r\n" ); document.write( " * 123° *\r\n" ); document.write( " ******************************\r\n" ); document.write( " 7\r\n" ); document.write( "\r \n" ); document.write( "\n" ); document.write( "After 30 minutes (0.5 hours), the ship that has a speed of 10 mph has traveled 5 miles and the ship that has a speed of 14 mph has traveled 7 miles. Given is that the two ships are 123° apart. So, this all forms a triangle. (See above.) We are looking for C. \n" ); document.write( " \n" ); document.write( "Use the law of cosines to find C: \n" ); document.write( "c² = a² + b² - 2ab(cos c) \n" ); document.write( " \n" ); document.write( "Using our numbers: \n" ); document.write( "c² = 5² + 7² - 2(5)(7)(cos 123°) \n" ); document.write( " \n" ); document.write( "c² = 25 + 49 - 70(cos 123°) \n" ); document.write( " \n" ); document.write( "c² = 74 - 70(cos 123°) \n" ); document.write( " \n" ); document.write( "c² = 74 - 70(-0.54464) \n" ); document.write( " \n" ); document.write( "c² = 74 + 38.1248 \n" ); document.write( " \n" ); document.write( "c² = 112.1248 \n" ); document.write( " \n" ); document.write( "c = 10.5889 \n" ); document.write( " \n" ); document.write( "The two ships are 10.5889 miles apart after 30 minutes. |