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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Algebra.Com's Answer #753913 by VFBundy(438)\"\" \"About 
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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" );
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\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.
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\n" ); document.write( "Use the law of cosines to find C:
\n" ); document.write( "c² = a² + b² - 2ab(cos c)
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\n" ); document.write( "Using our numbers:
\n" ); document.write( "c² = 5² + 7² - 2(5)(7)(cos 123°)
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\n" ); document.write( "c² = 25 + 49 - 70(cos 123°)
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\n" ); document.write( "c² = 74 - 70(cos 123°)
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\n" ); document.write( "c² = 74 - 70(-0.54464)
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\n" ); document.write( "c² = 74 + 38.1248
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\n" ); document.write( "c² = 112.1248
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\n" ); document.write( "c = 10.5889
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\n" ); document.write( "The two ships are 10.5889 miles apart after 30 minutes.
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