document.write( "Question 1130256: Thank you in advance for your help! I was really struggling with this problem:\r
\n" ); document.write( "\n" ); document.write( "Two planes take off from the same airport at the same time using different runways. One plane travels on a bearing S 15 degrees W at 250 miles per hour. The other plane travels on a bearing
\n" ); document.write( " N 75 degrees E at 300 miles per hour. How far are the planes from each other 2 hours after​ takeoff?\r
\n" ); document.write( "\n" ); document.write( "Answer: The distance is approximately ___ miles ( round to the nearest mile as needed)\r
\n" ); document.write( "\n" ); document.write( "THANK YOU AGAIN FOR YOUR HELP!!
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Algebra.Com's Answer #746940 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
my diagram is shown below.
\n" ); document.write( "once you have figured out how to draw the diagram, the rest is application of the law of cosines for angle C.\r
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\n" ); document.write( "\n" ); document.write( "that law is:\r
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\n" ); document.write( "\n" ); document.write( "c^2 = a^2 + b^2 - 2 * a * b * cos(C)\r
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\n" ); document.write( "\n" ); document.write( "side b is equal to 2 * 250 = 500 miles.
\n" ); document.write( "side a is equal to 2 * 300 = 600 miles.\r
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\n" ); document.write( "\n" ); document.write( "angle ACS is 15 degrees west of south.\r
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\n" ); document.write( "\n" ); document.write( "angle NCB is 75 degrees east of north.\r
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\n" ); document.write( "\n" ); document.write( "since angle NCE is 90 degrees, angle BCE is 15 degrees.\r
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\n" ); document.write( "\n" ); document.write( "angle SCE is 90 degrees.\r
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\n" ); document.write( "\n" ); document.write( "therefore angle ACB is equal to 15 + 90 + 15 = 120 degrees.\r
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\n" ); document.write( "\n" ); document.write( "the rest is application of the law of cosines to get c = 954 miles.\r
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\n" ); document.write( "\n" ); document.write( "that's the distance between the 2 planes after 2 hours.\r
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