document.write( "Question 233386: A car travels along a straight road, heading east for 1 h, then traveling for 30 min on another road that leads northeast. If the car has maintained a constant speed of 40 mi/h, how far is it from its starting position? Round your answer to 2 decimal places. \n" ); document.write( "
Algebra.Com's Answer #172275 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
After traveling an hour at 40mph the car traveled 40 miles. And after another 30 minutes it traveled another 20 miles.
\n" ); document.write( "Picture (or graph) the following:
  • Let's make the starting point the origin (point O), with east being to the right (along the x-axis).
  • After the first hour the car is at the point (40, 0). Let's call this point A.
  • The car then starts traveling northeast. A line in this direction would make a 45 degree angle with the x-axis.
  • After traveling 20 miles (30 minutes), we reach another point which we'll call B.
  • In the triangle formed by O, A and B:
    • OA = 40
    • AB = 20
    • angle OAB = 135 degrees
    • OB is the distance between the starting position and ending position.

\n" ); document.write( "Once we get this above, we should see that this is a problem made for the Law of Cosines: \"c%5E2+=+a%5E2+%2B+b%5E2+-+2ab%2Acos%28C%29\". Using this on your problem we get:
\n" ); document.write( "\"%28OB%29%5E2+=+%28OA%29%5E2+%2B+%28AB%29%5E2+-+2%28OA%29%28AB%29%2Acos%28OAB%29\"
\n" ); document.write( "Substituting our numbers we get:
\n" ); document.write( "\"%28OB%29%5E2+=+%2840%29%5E2+%2B+%2820%29%5E2+-+2%2840%29%2820%29%2Acos%28135%29\"
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"%28OB%29%5E2+=+1600+%2B+400+-+1600%2A%28%28-sqrt%282%29%29%2F2%29\"
\n" ); document.write( "\"%28OB%29%5E2+=+2000+%2B+800sqrt%282%29\"
\n" ); document.write( "Find the square root of each side:
\n" ); document.write( "\"sqrt%28%28OB%29%5E2%29+=+sqrt%282000+%2B+800sqrt%282%29%29\"
\n" ); document.write( "\"OB+=+sqrt%282000+%2B+800sqrt%282%29%29\"
\n" ); document.write( "This is the exact distance. I'll leave it up to you to use your calculator to come up with a decimal approximation, rounded to the nearest hundredth. \n" ); document.write( "
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