Component vectors positive or negative and angles positive or negative

[Who,me?]

Homework Statement

A cruise ship leaving port, travels 50.0 km 45.0 degrees north of west and then 70.0 km 30 degrees north of east. Find
a. The ship's displacement vector (the answer is Rx=25.3 km, Ry=70.4)
b. The displacement vector's magnitude and direction (the answer is 74.8 km 70.2 degrees north of east)

Homework Equations

SOH CAH TOA in these forms
Component vectors (x and y)
X cos(angle)
X sin (angle)
Then Pythagorean equation
asquared+bsquared=csquared

The Attempt at a Solution

Here's what I am doing;
I draw a coordinate plane, my first vector is drawn 45 degrees north from the x plane, I then draw the vector going 50 km north of west into the 2nd quadrant. From that point I draw a straight line directly right as a reference point for east. I draw a 30 angle from that straight line and begin drawing the 70 km vector.
Now,
Here's what I need; the displacement vector is a straight line from my starting point directly to my finishing point. However, since it is not a right triangle I will need to find x and y component vectors for both my first vector (50km) and my second vector (70km). After that I will need to add up both x component vectors and then separately add up both y component vectors. These new components are my Rx and Ry i.e. my x and y components of my displacement vector. I take Rx squared plus Ry squared and then find the square root of that. This should be my displacement vector, but it isn't working!

Here are somethings that might help:
Are all the angles I am going to put into my equations positive? If not, which ones are negative and why?
Are all my vectors positive here? It's hard to tell the way I placed things on the coordinate plane. Which vectors are supposed to be negative, it seems to me that the x component of B (the second 70km one) would be negative is that right?
Am I going about this the right way? I need to break everything down into right triangles right?

 

[nickjer]

Which direction is west? North of west doesn't usually mean a vector north of the positive x-axis. But this will be real hard to judge anything unless you actually post your picture here.

 

[Who,me?]

I don't think you can post attachments on this thing, or even photos.

 

[Squeezebox]

Towards the left of the y-axis is -x direction and right of the y-axis is +x direction. So the x-component of one of the vectors is negative. Which one is it?

 

[nickjer]

You can post pictures here.

 

[dwaynem]

Using + = east and north of origin:
Total E-W displacement = (-)50 cos45 + 70 cos 30 = 25.266East of origin
total N-S displacement = 50 sin45 + 70 sin 30 = 70.355 north of origin

Since you're talking in terms of north-south and east-west of origin, you do have right triangles to deal with:
length of resultant = sqrt(25.266² + 70.355²) = 74.754
angle north of east = tan(-1) (70.355/25.266) = 70.24 degrees

physics + dirt = engineering