Question

Vector A is 2.80cm long and is 60° above the x-axis in the first quadrant. Vector B is 1.90cm long and is 60° below the x-axis in the fourth quadrant. Find the magnitude and direction of A Vector +B Vector
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Answer 1

Answer:

Ax = 2.80cos60° = 1.40 cm; Ay = 2.80sin60° = 2.42 cm

Bx = 1.90cos(-60°) = 0.95 cm, By = 1.90sin(-60°) = - 1.65 cm

So, (A - B)x = Ax - Bx = 1.40 - 0.95 = 0.45 cm

(A - B)y = Ay - By = 2.42 - (- 1.65) = 4.07 cm.

Then magnitude of (A - B) = (0.452 + 4.072)1/2 = 4.095 cm.

Because (A - B)x > 0 and (A - B)y > 0 this vector lokated in 1st quadrant.

Now tan∝ = 4.07/0.45 = 9.0444, so ∝ tan-1(9.0444) = 83.69°,

∝ is angle above x-axis