The angle between the vector and the positive x-axis will be greater than 90° in each case

Question

ii. Suppose that vector 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

To find the angle of a vector, you will need to use the tan^-1 function of the x and y components of the vector.  V is the distance of the vector.

 

           If

 

tanθ = sinθ / cosθ        

 

          and

 

x = V*cosθ 

y = V*sinθ

 

          then

 

tanθ = (y/x)  

 

θ = tan^-1(y/x)

 

Remember that these functions are negative or positive depending on the quadrants they lie on.

 

Sinθ is only positive in Quadrant 1 and 2. Therfore, the positive angle in Quardrant 2  (90° ≤ θ ≤ 180°) is

 

180 - tan^-1(y/x)

 

 

Tanθ is only positive in Quadrant 1 and 3. Therefore, the positive angle in Quadrant 3 (180° ≤ θ ≤ 270°) is

 

180 + tan^-1(y/x)

 

 

Lastly, Cosθ is only positive in Quadrant 1 and 4.  Therefore the positive angle in Quadrant 4 (270° ≤ θ ≤ 360°) is

 

360 - tan^-1(y/x)

 

 

You now have your positive angle in their respective quadrants.

Answer 2

For "r" = magnitude of the vector <x, y>

 

sin^-1 y/r = θ

cos^-1 x/r = θ