Vector problem(A+B=C) A and C given

[Frenchie]

Homework Statement

In the sum A+B = C, vector A has a magnitude of 10.6 m and is angled 41.3° counterclockwise from the +x direction, and vector C has a magnitude of 16.5 m and is angled 13.1° counterclockwise from the -x direction. What are (a) the magnitude and (b) the angle (relative to +x) of B ? State your angle as a positive number.

Homework Equations

The Attempt at a Solution

I found the components of vector A and C which came out to be

Ay=10.6*sin(41.3)=6.99601768
Ax10.6*cos(41.3)=7.963399815

Cy=16.5*sin(13.1)=3.739746573
Cx=16.5*cos(13.1)=16.0760346

so I'm not sure what i do after that. Do i add the components or subtract them to get By and Bx and then get the square root of both y and x squared? Thank you for any help given!~

 

[mighty2000]

Hello,

Thank you for your question. To find the magnitude and angle of vector B, you can use the Pythagorean theorem and trigonometric functions.

(a) The magnitude of vector B can be found by using the Pythagorean theorem: B = √(Bx^2 + By^2). To find the components of vector B, you can use the fact that the sum of the components of A, B, and C must equal zero. So, Bx = -Ax - Cx and By = -Ay - Cy. Substituting the values you calculated for Ax, Ay, Cx, and Cy, you can find the components of B and then use the Pythagorean theorem to find the magnitude.

(b) To find the angle of vector B, you can use the trigonometric function tangent. The tangent of an angle is equal to the ratio of the opposite side to the adjacent side. In this case, the angle you are looking for is the angle between vector B and the +x direction, so you can use the components of B to find this angle. The angle can be found by taking the inverse tangent (tan^-1) of By/Bx. Make sure to take the absolute value of this angle, as the inverse tangent function will give you a negative angle. This will give you the angle in radians. To convert it to degrees, you can multiply it by 180/π.

I hope this helps. Let me know if you have any further questions.