Finding the Magnitude of Vector Sum from Components

[Want to learn]

Homework Statement

Vector C has a magnitude of 5.0 units and makes an angle of -90.0º with the positive x-axis, vector D has a magnitude of 7.0 units and makes and angle of –120º with the positive x-axis. What is the magnitude of the vector sum of C + D? - I am assuming that this means the resultant.

Homework Equations

General for finding the components:

$$A_x = Acos\theta$$

$$A_y = Asin\theta$$

Magnitude:

$$A = \sqrt {{A_x}^2 + {A_y}^2}$$

The Attempt at a Solution

I first start with $$\vec C$$

$$C_x = 5.0 cos (-90) = 0$$
$$C_y = 5.0 sin (-90) = -5$$

Move on to $$\vec D$$

$$D_x = 7.0 cos (-120) = -3.5$$
$$D_y = 7.0 sin (-120) = -6.1$$

I have all the components now, moving on to finding the A + B - which I am assuming means the resultant.

$$R_x = 0 + (-3.5) = -3.5$$
$$R_y = -5 + (-6.1) = -11.1$$

$$R = \sqrt { (-3.5)^2 + (-11.1)^2} = 11.6$$ - This is my answer to the question

Now, I am not sure if my assumption that A+B = Resultant is true. Any ideas on where I am messing up. Thank you

 

[Saladsamurai]

Are you sure that you copied this correctly? If you were given C and D I don't see why they would be asking you for A+B.

 

[Want to learn]

Are you sure that you copied this correctly? If you were given C and D I don't see why they would be asking you for A+B.

Sorry m8, had a different problem in my head. I made the change.

Any ideas on what I did wrong?

 

[Saladsamurai]

Sorry m8, had a different problem in my head. I made the change.

Any ideas on what I did wrong?

Not sure. It looks good to me

Why do you think you are wrong? Did you copy the numbers down correctly?

I get the same answer using the numbers you gave.

 

[Want to learn]

Ok well if that looks right, then great! Now there is a second part to this question which I did not post and it states the same variables and measurements except you have to find the direction of the vector sum C+D referenced to the positive x-axis.

First I use this equation:

$$tan \theta = \frac {A_y} {A_x}$$

So...

so I solve for theta and get:

$$\theta = tan^-1 \frac {A_y} {A_x}$$

So...

$$\theta = tan^-1 \frac {-11.5} {-3.5} = 73.07...$$

Looks like the vector is in the third quadrant so I add 180 to $$\theta$$ and get 253.1. The problem is that none of the above answers are part of my answer choices. What did I do wrong?

 

[rl.bhat]

Ry is 11.1 not 11.5

 

[Want to learn]

yah another typo. That still doesn't make a difference, I still have it wrong.

 

[rl.bhat]

yah another typo. That still doesn't make a difference, I still have it wrong.

What are the answer choices?