Friday

August 22, 2014

August 22, 2014

Posted by **Sam** on Sunday, June 15, 2008 at 2:54pm.

- Vectors -
**bobpursley**, Sunday, June 15, 2008 at 4:18pmUse the law of cosines: You know two sides, and the included angle.

Draw the figure before you compute.

- Vectors -
**Sam**, Sunday, June 15, 2008 at 6:14pmI tried that but it doesn't work.

I did (5)(2^2)+(2)(3^2)-(2)(2)(3)cos50.

- Vectors -
- Vectors -
**Damon**, Sunday, June 15, 2008 at 9:51pmvector 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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