What should be the angle between two vectors of magnitudes 4.10 and 4.40 units, so that their resultant has a magnitude of 3.90 units?
To find the angle between two vectors, you can use the cosine rule. The cosine rule states that in a triangle with sides A, B, and C, and angle θ opposite side C, the following equation holds:
C^2 = A^2 + B^2 - 2AB*cos(θ)
In this case, the magnitude of the first vector is 4.10 units (let's call it A), the magnitude of the second vector is 4.40 units (let's call it B), and the magnitude of their resultant is 3.90 units (let's call it C).
So, we have:
C^2 = A^2 + B^2 - 2AB*cos(θ)
(3.90)^2 = (4.10)^2 + (4.40)^2 - 2 * 4.10 * 4.40 * cos(θ)
Simplifying the equation:
15.21 = 16.81 + 19.36 - 35.944 * cos(θ)
15.21 = 36.17 - 35.944 * cos(θ)
Rearranging the equation:
35.944 * cos(θ) = 36.17 - 15.21
35.944 * cos(θ) = 20.96
Dividing both sides of the equation by 35.944:
cos(θ) = 20.96 / 35.944
cos(θ) ≈ 0.583001
Now, to find the angle θ, we need to take the inverse cosine (cos^(-1)) of 0.583001:
θ = cos^(-1)(0.583001)
θ ≈ 54.11 degrees
Therefore, the angle between the two vectors should be approximately 54.11 degrees.