The magnitudes of two vectors are $8$ units and $6$ units respectively. Find the magnitude of the resultant vector if the angle between these two vectors is $(i) \theta = 0^{\circ}$,$(ii) \theta = 180^{\circ}$,$(iii) \theta = 90^{\circ}$,and $(iv) \theta = 120^{\circ}$.

(N/A) The magnitude of the resultant vector $R$ of two vectors $A$ and $B$ with an angle $\theta$ between them is given by $R = \sqrt{A^2 + B^2 + 2AB \cos \theta}$.
$(i)$ For $\theta = 0^{\circ}$,$\cos 0^{\circ} = 1$.
$R = \sqrt{8^2 + 6^2 + 2(8)(6)(1)} = \sqrt{64 + 36 + 96} = \sqrt{196} = 14$ units.
$(ii)$ For $\theta = 180^{\circ}$,$\cos 180^{\circ} = -1$.
$R = \sqrt{8^2 + 6^2 + 2(8)(6)(-1)} = \sqrt{64 + 36 - 96} = \sqrt{4} = 2$ units.
$(iii)$ For $\theta = 90^{\circ}$,$\cos 90^{\circ} = 0$.
$R = \sqrt{8^2 + 6^2 + 2(8)(6)(0)} = \sqrt{64 + 36} = \sqrt{100} = 10$ units.
$(iv)$ For $\theta = 120^{\circ}$,$\cos 120^{\circ} = -0.5$.
$R = \sqrt{8^2 + 6^2 + 2(8)(6)(-0.5)} = \sqrt{64 + 36 - 48} = \sqrt{52} = 2\sqrt{13}$ units.