Find the angle between 2n force and 3n force so their resultant is 4n
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Let the angle be ¢
√((2n^2)+(3n^2)+2*(2n)(3n)cos¢)=4n
4(n^2)+9(n^2)+24(n^2)(cos¢) =16(n^2)
24(n^2)(cos¢)=3(n^2)
1/8=cos¢
so the angle becomes approximately 83°
√((2n^2)+(3n^2)+2*(2n)(3n)cos¢)=4n
4(n^2)+9(n^2)+24(n^2)(cos¢) =16(n^2)
24(n^2)(cos¢)=3(n^2)
1/8=cos¢
so the angle becomes approximately 83°
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