Two vectors A and B have equal magnitudes. The magnitude of $(\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}})$ is ‘n’ times the magnitude of $(\overrightarrow{A}-\overrightarrow{B})$ The angle between $\overrightarrow{A}$ and $\overrightarrow{B}$ is

$$\begin{gathered} \mathrm{A}=\mathrm{B}=\mathrm{x} \\ |\overline{\mathrm{A}}+\overline{\mathrm{B}}|=\mathrm{n}|\overline{\mathrm{A}}-\overline{\mathrm{B}}| \end{gathered}$$

${\mathrm{A}}^2+{\mathrm{B}}^2+2\mathrm{ABcos}\mathrm{\theta }={\mathrm{n}}^2\left[{\mathrm{A}}^2+{\mathrm{B}}^2-2\mathrm{ABcos}\mathrm{\theta }\right]$

$2{\mathrm{x}}^2+2{\mathrm{x}}^2\cos \mathrm{\theta }={\mathrm{n}}^2\left[2{\mathrm{x}}^2-2{\mathrm{x}}^2\cos \mathrm{\theta }\right]$

$2{\mathrm{x}}^2[1+\cos \mathrm{\theta }]={\mathrm{n}}^{2}2{\mathrm{x}}^2[1-\cos \mathrm{\theta }]$

$1+\cos \theta ={\mathrm{n}}^2-{\mathrm{n}}^2\cos \theta$

$\cos \mathrm{\theta }\left[{\mathrm{n}}^2+1\right]=\left[{\mathrm{n}}^2-1\right]\Rightarrow \mathrm{\theta }={\cos }^{-1}\left[\frac{{\mathrm{n}}^2-1}{{\mathrm{n}}^2+1}\right]$