In the sum Vector A plus Vector B equals vector C, vector A has a magnitude of 11.6 m and is angled 36.5° counterclockwise from the +x direction, and vector C has a magnitude of 14.0 m and is angled 17.1° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of Vector B.

You must have meant,

"What are (a) the magnitude and (b) the angle (relative to +x) of Vector C "

A = (11.6cos36.5 + 11.6sin36.5) = (appr 9.32, 6.90)
B = (14cos17.1 + 14sin17.1) = (appr13.38 , 4.12)

vector C = (22.71 , 11.02)

magnitude = √22.71^2 + 11.02^2) = 25.24
angle:
tanØ = 11.02/22.71
Ø = 25.88°

To find the magnitude and angle of Vector B, we can use vector addition and basic trigonometry.

(a) Magnitude of Vector B:
We know that the sum of Vector A and Vector B equals Vector C. In vector addition, the x-components and y-components of the vectors add up separately.

The x-component of Vector A is given by:
Ax = A * cos(θA)
where A is the magnitude of Vector A and θA is the angle of Vector A.

Similarly, the x-component of Vector B is given by:
Bx = B * cos(θB)
where B is the magnitude of Vector B and θB is the angle of Vector B.

The total x-component of Vector C can be calculated as the sum of the x-components of Vector A and Vector B:
Cx = Ax + Bx

Since we know the magnitudes and angles of Vector A and Vector C, we can substitute them into the equations to solve for B.

1. Calculate the x-components:
Ax = A * cos(θA) = 11.6 m * cos(36.5°)
Cx = C * cos(θC) = 14.0 m * cos(17.1°)

2. Substitute the values:
Cx = Ax + Bx
14.0 m * cos(17.1°) = 11.6 m * cos(36.5°) + B * cos(θB)

3. Solve for B:
B * cos(θB) = 14.0 m * cos(17.1°) - 11.6 m * cos(36.5°)
B * cos(θB) = (14.0 m * cos(17.1°)) - (11.6 m * cos(36.5°))
B = [(14.0 m * cos(17.1°)) - (11.6 m * cos(36.5°))] / cos(θB)

Now you can substitute the values you have to get the magnitude of Vector B.

(b) Angle of Vector B:
To find the angle of Vector B, we will use the y-components of the vectors.

The y-component of Vector A is given by:
Ay = A * sin(θA)

Similarly, the y-component of Vector B is given by:
By = B * sin(θB)

The total y-component of Vector C can be calculated as the sum of the y-components of Vector A and Vector B:
Cy = Ay + By

Since we know the magnitudes and angles of Vector A and Vector C, we can substitute them into the equations to solve for θB.

1. Calculate the y-components:
Ay = A * sin(θA) = 11.6 m * sin(36.5°)
Cy = C * sin(θC) = 14.0 m * sin(17.1°)

2. Substitute the values:
Cy = Ay + By
14.0 m * sin(17.1°) = 11.6 m * sin(36.5°) + B * sin(θB)

3. Solve for θB:
B * sin(θB) = 14.0 m * sin(17.1°) - 11.6 m * sin(36.5°)
θB = arcsin([(14.0 m * sin(17.1°)) - (11.6 m * sin(36.5°))]) / B

Now you can substitute the values you have to find the angle of Vector B relative to the +x direction.