document.write( "Question 483951: Ajax is 8 km due west of Oshawa. Uxbridge is 16 km NW of Oshawa. How far is it from Ajax to Uxbridge? Explain whether you have enough information to solve this problem.\r
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\n" ); document.write( "\n" ); document.write( "my answer was 13.856 km but not confident how I found the angle oshawa to begin with. I used Cos Oshawa = a/h = 8/16, cos-1 0.5=60 degrees. \r
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Algebra.Com's Answer #331240 by ankor@dixie-net.com(22740)\"\" \"About 
You can put this solution on YOUR website!
Ajax is 8 km due west of Oshawa. Uxbridge is 16 km NW of Oshawa.
\n" ); document.write( "How far is it from Ajax to Uxbridge?
\n" ); document.write( " Explain whether you have enough information to solve this problem.
\n" ); document.write( ":
\n" ); document.write( "This does not form a right triangle, however we know that the angle at Oshawa is 45 degrees, (NW with W forms a 45 degrees)
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\n" ); document.write( "Let this angle be A, then a = the dist from Ajax to Uxbridge
\n" ); document.write( "Using the law of cosine to find a
\n" ); document.write( "a^2 = b^2 + c^2 - 2bc*cos(A)
\n" ); document.write( "where
\n" ); document.write( "a = dist from Ajax to Uxbridge
\n" ); document.write( "b=8
\n" ); document.write( "c=16
\n" ); document.write( "A=45
\n" ); document.write( ":
\n" ); document.write( "a^2 = 8^2 + 16^2 - 2(8*16)*cos(45)
\n" ); document.write( "a^2 = 64 + 256 - 2(128)*.707
\n" ); document.write( "a^2 = 320 - 256*.707
\n" ); document.write( "a^2 = 320 - 181
\n" ); document.write( "a = \"sqrt%28139%29\"
\n" ); document.write( "a = 11.79 km
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