document.write( "Question 328202: Jan is flying on a triangular course at 320 mi/h. She flies due east for two hours and then turns right through a 65 degree angle. How long after turning will she be exactly southeast of where she started?
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Algebra.Com's Answer #235068 by galactus(183)\"\" \"About 
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There are various ways to go about this one. Here is one method.\r
\n" ); document.write( "\n" ); document.write( "From where she makes her 65 degree turn, extend a line straight out, then connect it to the point where she intersects the southwest line. \r
\n" ); document.write( "\n" ); document.write( "This forms a right triangle. Draw a line from the intersection point back to where she turned. This forms another triangle.\r
\n" ); document.write( "\n" ); document.write( "Let the distance extended out from where she turned be x and the distance from there to the intersection point be 640+x (because southeast is 45 degrees from the origin)\r
\n" ); document.write( "\n" ); document.write( "The distance she flies is 320t.\r
\n" ); document.write( "\n" ); document.write( "Now, we can build two triangles and solve for t and x. t is what we really need.\r
\n" ); document.write( "\n" ); document.write( "\"sin%2825%29=x%2F320t\"\r
\n" ); document.write( "\n" ); document.write( "\"x%5E2%2B%28640%2Bx%29%5E2=%28320t%29%5E2\"....[2]\r
\n" ); document.write( "\n" ); document.write( "\"x=320t%2Asin%2825%29\"....[3]\r
\n" ); document.write( "\n" ); document.write( "Sub [3] into [2] and solve for t.\r
\n" ); document.write( "\n" ); document.write( "\"%28320t%2Asin%2825%29%29%5E2%2B%28640%2B%28320t%2Asin%2825%29%29%29%5E2=%28320t%29%5E2\"\r
\n" ); document.write( "\n" ); document.write( "\"204800sin%5E2%2825%29%2At%5E2%2B409600sin%2825%29t%2B409600=102400t%5E2\"\r
\n" ); document.write( "\n" ); document.write( "This is a quadratic to solve for t.\r
\n" ); document.write( "\n" ); document.write( "Doing so, gives us \"t=4.13\" hours.\r
\n" ); document.write( "\n" ); document.write( "The other solution is extraneous.
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