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  68.79 N, 13.84° N of W

Explanation:

The law of cosines can be used to find the magnitude of the sum. F1 is 30° N of W, and F2 is 30° S of W, so the exterior angle of the force triangle is 30°+20° = 50°. The interior angle is the supplement of that. The angle between F1 and F2 in the force triangle representing the sum is 130°, so the sum of forces is ...

  |F|^2 = |F1|^2 +|F2|^2 -2·|F1|·|F2|·cos(130°)

  = 50^2 +25^2 -2·50·25·cos(130°) ≈ 4731.969

  |F| ≈ √4731.969 ≈ 68.79 . . . . newtons

The angle α between F and F1 can be found from the law of sines.

  sin(α)/|F2| = sin(130°)/|F|

  α = arcsin(|F2|/|F|·sin(130°)) ≈ 16.16°

The diagram shows this to be the angle south of F1, so the angle of the sum vector F is 30° -16.16° N of W = 13.84° N of W.

The resultant force vector is 68.79 N at an angle of 13.84° N of W.

68.98N, 13.8° N of W

Explanation:

The Forces F1+ F2 = 50N 30°N of W + 25N 20°S of W.

This forces can be split into horizontal and vertical components and are sum as such.

The horizontal and vertical component of F1 are;

50N cos 30= 43.30N W

50N sin 30 = 25N

The horizontal and vertical component of F2 are;

25N cos20°=23.49N West

25N sin20°=8.55N South

Sum of horizontal forces =43.30N+23.49 = 66.99N

Sum of vertical forces =25-8.55=16.45N{North is at the positive side of the y axis and South at the negative side}

The resultant sum of this forces is √(66.99)^2 + (16.45)^2=√4758.26=68.98N

The angle at which this force moves is Tan^{-1} 16.45/66.99 = 13.8° N of W

The Force therefore is 68.98N, 13.8° N of W

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