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)
SOH CAH TOA in these forms
Component vectors (x and y)
X sin (angle)
Then Pythagorean equation
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.
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?