Interact with the app below to discover and apply the formulas to find the components, the magnitude and the angle of a vector.[br][br]A note below the app explains how to find the correct value of [math]\theta[/math] using a calculator.

The [b]angle [/b][math]\theta[/math] that a vector [b]v[/b] forms with the positive [i]x[/i]-axis describes its [b]direction[/b] in a two-dimensional Cartesian coordinate system.[br][br]We know that [math]\tan\theta=\frac{v_y}{v_x}[/math]. [br]If we use the [math]\tan^{-1}[/math] key of our calculator, we always obtain a value for [math]\theta[/math] in the interval [math]\left(-90°,90°\right)[/math] or [math]\left(-\frac{\pi}{2},\frac{\pi}{2}\right)[/math] because that is the main interval in which the tangent function is invertible.[br][br]In the app above, drag point [math]B[/math] at [math]\left(-6,6\right)[/math], and assume that you know only the [i]x[/i]- and[i] y[/i]-components of the vector.[br]Applying the formula, you get [math]\tan\theta=\frac{6}{-6}=-1[/math], and using the [math]\tan^{-1}[/math] key of your calculator you get [math]-45°[/math] or [math]-\frac{\pi}{4}[/math], that are not the angle of [math]135°[/math] that you see in the app.[br][br]Because of the constraints on the result explained above, you need to [b]adjust the calculator result for the quadrant[/b], and precisely:[br][list][*]If [b]v[/b] is in the first quadrant [math]\left(v_x>0,v_y>0\right)[/math], the value of [math]\theta[/math] is correct as is[/*][*]If [b]v[/b] is in the second or third quadrant [math]\left(v_x<0\right)[/math], add [math]180°[/math] (or [math]\pi[/math] radians) to ensure [math]\theta[/math] in the 2nd or 3rd quadrant[/*][*]If [b]v[/b] is in the 4th quadrant [math]\left(v_x>0,v_y<0\right)[/math], add [math]360°[/math] (or [math]2\pi[/math] radians) to ensure [math]\theta[/math] is in the 4th quadrant.[br][/*][/list][br]