What is the angle between a vector and the positive x-axis?

[mbecker19]

The angle (in degrees) between A = -24.5i + 45.4j, and the positive x-axis is?

 

[rl.bhat]

If A = xi + yj, then tan(theta) = y/x

 

[mbecker19]

i still don't know how to figure out the angle I am confused
i know the magnitude of the vector is 38.3 but what is the x or y component

 

[mbecker19]

actually i figured it out all i had to do was set tan-1(45.4/-24.5) and got 61.65 and then added the 28.35 which was the extra onto the already existing 90 degrees and got 118.35 deg
thank you

 

[rl.bhat]

In this vector, x component is -24.5 and y component is 45.4. To find the angle the magnitude of the vector in not needed.
And the magnitude of the vector is given by (x^2 + y^2)^1/2

 

[rl.bhat]

You have to wright it as -118.35 deg, because according to convention, if you measure the angle from the x-axis in the antyclockwise direction, the angle is positive. Otherwise it is negative.