At what angle two vectors of magnitudes $\text{A}+\text{B}\text{and}\text{A}-\text{B}$ act, so that their resultant is $\sqrt{3{\text{A}}^{\text{2}}+{\text{B}}^{\text{2}}}$

$3A^2+B^2=2A^2+2B^2+2\left(A^2-B^2\right)\cos \theta$

$\left(A^2-B^2\right)-2\left(A^2-B^2\right)\cos \theta =0$

$\left(A^2-B^2\right)-\left(1-2\cos \theta \right)=0$

$\therefore \left(1-2\cos \theta \right)=0$

$\cos \theta =\frac{1}{2}$

$\theta =60°$