The resultant of two vectors of magnitudes $\left(\text{P}+\text{Q}\right)\text{and}\left(\text{P}-\text{Q}\right)\text{is}\sqrt{\left({\text{P}}^{\text{2}}+{\text{Q}}^{\text{2}}\right)}$ . Then the angle between the vectors is

$\text{R}=\sqrt{{\text{P}}^2+{\text{Q}}^{\text{2}}+2\text{PQ}\cos \theta }$

$\left|\overline{P}\right|=\text{P}+\text{Q}$

$\left|\overline{Q}\right|=\text{P}-\text{Q}$

${\text{P}}^2+{\text{Q}}^2={\left(\text{P}+\text{Q}\right)}^2+{\left(\text{P}-\text{Q}\right)}^2+2\left(\text{P}+\text{Q}\right)\left(\text{P}-\text{Q}\right)\cos \theta$

${\text{P}}^2+{\text{Q}}^2={\text{P}}^2+{\text{Q}}^2+2\text{PQ}+{\text{P}}^2+{\text{Q}}^2-2\text{PQ}+2\left({\text{P}}^2+{\text{Q}}^2\right)$

$=2{\text{P}}^2-2{\text{Q}}^2+2\left({\text{P}}^2-{\text{Q}}^2\right)\cos \theta$

$-\left({\text{P}}^2+{\text{Q}}^2\right)=2\left({\text{P}}^2+{\text{Q}}^2\right)\cos \theta$

$\cos \theta =\frac{-\left({\text{P}}^2+{\text{Q}}^2\right)}{2\left({\text{P}}^2-{\text{Q}}^2\right)}$

$\theta ={\cos }^{-1}\left(\frac{-\left({\text{P}}^2+{\text{Q}}^2\right)}{2\left({\text{P}}^2-{\text{Q}}^2\right)}\right)$