Question

Suppose that vector v lies in quadrant II, quadrant III, or quadrant IV. How can you use trigonometric ratios to calculate the direction (i.e., angle) of the vector in each of these quadrants with respect to the positive x-axis? The angle between the vector and the positive x-axis will be greater than 90 in each case.

Answer 1

I just had a test on vectors so I feel your pain! Explained in the picture above. Hope it helps C:
Edit: in picture I say tantheta = x/y sorry I meant y/x

Answer 2

The angles with respect the positive x-axis are given by the following formulas for each quadrant.

Second quadrant

`\theta =\tan^(-1) \left(\cfrac yx\right)+\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+180^\circ`

Third quadrant

`\theta =\tan^(-1) \left(\cfrac yx\right)+\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+180^\circ`

Fourth quadrant

`\theta =\tan^(-1) \left(\cfrac yx\right)+2\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+360^\circ`

Explanation:

In order to calculate the angle in a general way, we always use the formula

`\theta =tan^(-1) \left( \cfrac yx \right)`

That will give us the angle of the associated reference triangle, the issue is that the the output of the inverse trigonometric function is in the range
`[-\pi/2, \pi/2]` and that does not give us the exact angle with respect the positive x-axis.

Thus we need to add some angle to get the correct angle with respect the positive x-axis.

Case second quadrant.

The output of
`\theta =tan^(-1) \left( \cfrac yx \right)` will be negative, since y is greater than 0 but x is less than zero.

To fix that, and get the angle on the second quadrant we can add either pi or 180 degrees, so we will get:

`\theta =\tan^(-1) \left(\cfrac yx\right)+\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+180^\circ`

Case third quadrant.

The output of
`\theta =tan^(-1) \left( \cfrac yx \right)` will be positive but it will be smaller than 90 degrees.

To fix that, and get the angle on the second quadrant we can add either pi or 180 degrees, so we will get:

`\theta =\tan^(-1) \left(\cfrac yx\right)+\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+180^\circ`

Case fourth quadrant

The output of
`\theta =tan^(-1) \left( \cfrac yx \right)` will be negative since y is less than 0 but x is positive.

To fix that we can get a positive coterminal angle, that means adding 2pi or 360 degrees, so we will get

`\theta =\tan^(-1) \left(\cfrac yx\right)+2\pi\\\theta =\tan^(-1) \left(\cfrac yx\right)+360^\circ`