Solving for Vector Magnitude and Direction: Ax=3.2, Ay=-5.15 - Quick Question

[joe215]

Homework Statement

Find the vector with the components: Ax=3.2 Ay= -5.15

The Attempt at a Solution

-5.15^2 + 3.2^2= 36.76
sq root of 36.76= 6.06
So, the magnitude is 6.06. Now for the direction...

CosD=3.2/6.06
ArcCos(0.52)=58.7 degrees
So, the direction is 58.7

Am I right? Thanks so much!

 

[rootX]

Homework Statement

Find the vector with the components: Ax=3.2 Ay= -5.15

The Attempt at a Solution

-5.15^2 + 3.2^2= 36.76
sq root of 36.76= 6.06
So, the magnitude is 6.06. Now for the direction...

CosD=3.2/6.06
ArcCos(0.52)=58.7 degrees
So, the direction is 58.7

Am I right? Thanks so much!

with y negative and x positive, the vector falls in fourth quadrant. But 58.7 is in first..

other than that, everything else seems good.

try using tan theta = y/x .. it's easier to use

 

[joe215]

So does that mean...

Tan theta=-5.15/3.2
ArcTan(-1.6)= -57.99 degrees

How can I have a negative angle?