Two vectors $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ have precisely equal magnitudes. For the magnitude of $\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}$ to be larger than the magnitude of $\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}$ by the factor $n,$ what must be the angle between them?

Key Concepts: Vector Addition, Vector Subtraction, Magnitude Of Vectors Explanation: Let the magnitude of $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ be denoted as $a$ and angle between them as $\theta$. Using vector addition and subtraction rules we get $|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=\sqrt{a^2+a^2+2a^2\cos (\theta )}$ and $|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|=\sqrt{a^2+a^2-2a^2\cos (\theta )}$. We can now proceed to solve the problem by using these two equations and comparing them using the factor $n$. Step by Step Solution: Step 1. Compute the absolute value of the vector addition $|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=\sqrt{a^2+a^2+2a^2\cos (\theta )}$ Step 2. Compute the absolute value of the vector subtraction $|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|=\sqrt{a^2+a^2-2a^2\cos (\theta )}$ Step 3. Set up the equation $|\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{B}}|=n|\overrightarrow{\mathbf{A}}-\overrightarrow{\mathbf{B}}|$ and simplify Step 4. Substitute the previously calculated values to solve for $\theta$. Final Answer: $\theta ={\cos }^{-1}(\sqrt{(n+1)/2})$