Find the angle between two vector forces...?

draw a diagram of this sized as best you can

the two forces will be quite widely spread

now complete the parallelogram

let's look at the left side of the diagram

we have an isoscles triangle

just to make it easy le'ts call the vectors 3 and the diagonal 1

looking at the angle between the vector and the diagonal and calling it angle C

we can fit that into the Law of Cosines

c^2 = a^2 + b^2 - 2 a b cosC

3^2 = 3^2 + 1^2 - 2 (3) (1) cosC

9 = 9 + 1 - 6cosC

6cosC = 1

cosC = 1/6 = 0.1667

C = 80.4 deg

this is half the angle between the two vectors

so 2(80.4) = 160.8 degrees

From what you have, the resultant is less than the other sides of the triangle formed by the three vectors (the two forces and their resultant). Then the resultant is a difference, say, F=F1-F2. Such a configuration has been worked out already as the law of cosines. That is:

F^2 = F1^2 + F2^2 -2×F1×F2×cos(x) where x is the angle between F1 and F2. Since F1 = F2 and F = 1/3 F1, the equation above becomes:

(1/3 F1)^2 = 2×F1^2 -2×F1^2×cos(x) which reduces to

1/18 = 1 - cos(x). So you'll have cos(x) = 17/18 and hence x = arccos(17/18) = 19.188°.

initiate by ability of drawing your self a diagram | 12 lbs | | 60 ranges |._______20lbs____ considering that the two between the forces are vectors you could restructure them to type a parallelogram. for this reason, you could re-draw the form like this: |------------------------20lbs |one hundred twenty ranges* |12lbs | *The sum of adjoining angles in a parallelogram is a hundred and eighty ranges. by ability of drawing a resultant between your commencing element and your end element you presently have a triangle with an unknown element (x, the ensuing), 2 commonplace components and a commonplace perspective. From here you ought to use the regulation of Cosines: c^2 = a^2+b^2-2ab*cos(C) x^2 = 12^2 + 20^2 - 2*12*20*cos(one hundred twenty) x^2 = 784 x = 28 lbs.