document.write( "Question 43408: what is the magnitude
\n" ); document.write( "U: magnitude 1,850, bearing 125 degrees
\n" ); document.write( "V: magnitude 2,960, bearing 25 degrees
\n" ); document.write( "W: U+V \n" ); document.write( "
Algebra.Com's Answer #28432 by fractalier(6550)\"\" \"About 
You can put this solution on YOUR website!
The process of adding or subtracting vectors involving breaking the pieces down into their horizontal and vertical components first...then combining like components...then changing the components into polar form...for this one we have...I use q and p for theta and phi...
\n" ); document.write( "Ux = U cos q = 1850 cos 125° = -1061
\n" ); document.write( "Uy = U sin q = 1850 sin 125° = 1515
\n" ); document.write( "Vx = V cos p = 2960 cos 25° = 2683
\n" ); document.write( "Vy = V sin p = 2960 sin 25° = 1251
\n" ); document.write( "now add 'em
\n" ); document.write( "Wx = Ux + Vx = -1061 + 2683 = 1622
\n" ); document.write( "Wy = Uy + Vy = 1515 + 1251 = 2766
\n" ); document.write( "now
\n" ); document.write( "r = sqrt(x^2 + y^2) and theta = arctan(y/x) so that
\n" ); document.write( "r = sqrt(1622^2 + 2766^2) = 3206
\n" ); document.write( "and theta = arctan(2766/1622) = 59.6°
\n" ); document.write( "and finally (whew!)
\n" ); document.write( "W = magnitude 3206, bearing 59.6°
\n" ); document.write( "Study this example carefully because I probably won't do this kind of problem again... \n" ); document.write( "
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