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

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SUMMARY

The angle between the vector A = -24.5i + 45.4j and the positive x-axis is calculated using the formula tan(theta) = y/x. By substituting the values, tan-1(45.4/-24.5) yields an angle of 61.65 degrees. Adding 90 degrees results in an angle of 118.35 degrees, which should be expressed as -118.35 degrees due to the convention of measuring angles in the anticlockwise direction from the x-axis. The magnitude of the vector, calculated as (x^2 + y^2)^(1/2), is not necessary for determining the angle.

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The angle (in degrees) between A = -24.5i + 45.4j, and the positive x-axis is?
 
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If A = xi + yj, then tan(theta) = y/x
 
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
 
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
 
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
 
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.
 

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