Posted by Sam on Sunday, June 15, 2008 at 2:54pm.
Use the law of cosines: You know two sides, and the included angle.
Draw the figure before you compute.
I tried that but it doesn't work.
I did (5)(2^2)+(2)(3^2)-(2)(2)(3)cos50.
vector 5a has magnitude 10 and call it at 50 degrees
vector 2 b has magnitude 6 and call it at 0 degrees
so our triangle is
10 and 6 with fifty degrees between them
so the magnitude we are looking for is from
c^2 = 10^2 + 6^2 - 10*6*cos 50
that is why it did not work.
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