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December 18, 2014

December 18, 2014

Posted by **amaa** on Tuesday, December 4, 2012 at 7:33am.

- precalculus -
**Reiny**, Tuesday, December 4, 2012 at 8:10amlet AB be the original path, then veer off at 7° for BC.

Join AC

I see a long skinny triangle ABC with angle ABC = 173°

The lengths of AB and BC are 2.5(695) and 3(695) respectively, but we can just ignore the 695 for the time being and use the smaller similar triangle to keep the numbers low.

by cosine law

AC^2 = 2.5^2 + 3^2 - 2(2.5)(3)cos173

= 15.25 - (-14.88819...)

= 30.13819..

AC = 5.489826..

so the distance is 5.489..(695) = 3815.4 miles

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