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Q.
The resultant of two forces $(A+B)$ and $(A-B)$ is a force $\sqrt{3 A^{2}+B^{2}} .$ The angle between two given forces is
Motion in a Plane
Solution:
Here, $\vec{P}=(\vec{A}+\vec{B}), \vec{Q}=(\vec{A}-\vec{B}), \vec{R}$
$=\sqrt{3 A^{2}+B^{2}}$
Let $\theta$ be angle between the two given forces.
$\therefore \left(3 A^{2}+B^{2}\right)=(A+B)^{2}+(A-B)^{2}+2(A+B)(A-B) \cos \theta$
or $ 3 A^{2}+B^{2}=A^{2}+B^{2}+2 A B+A^{2}+B^{2}-2 A B+2\left(A^{2}-B^{2}\right) \cos \theta$
or $ A^{2}-B^{2}=2\left(A^{2}-B^{2}\right) \cos \theta$
or $\cos \theta=\frac{1}{2}=\cos \frac{\pi}{3}$ or
$\theta=\frac{\pi}{3}$