Q. If the resultant of two forces $(A + B)$ and $(A - B)$ is $\sqrt{A^{2} + B^{2}}$, then the angle between these forces is
A) $\cos^{- 1}\left\lbrack - \frac{(A^{2} - B^{2})}{A^{2} + B^{2}} \right\rbrack$
B) $\cos^{- 1}\left\lbrack - \frac{(A^{2} + B^{2})}{{(A}^{2} - B^{2})} \right\rbrack$
C) $\cos^{- 1}\left\lbrack - \frac{A^{2} + B^{2}}{2(A^{2} - B^{2})} \right\rbrack$
D) $\cos^{- 1}\left\lbrack - \frac{2(A^{2} + B^{2})}{A^{2} - B^{2}} \right\rbrack$
Explanation
(c) Here, $P = (A + B),\ \ Q = (A - B),$
$R = \sqrt{A^{2} + B^{2}};$
cos$\ \theta = \frac{R^{2} - P^{2} - Q^{2}}{2PQ}$
$= \frac{\left( A^{2} + B^{2} \right) - (A + B)^{2} - (A - B)^{2}}{2(A + B)(A - B)}$
$= - \left\lbrack \frac{A^{2} + B^{2}}{\begin{array}{r}
2(A^{2} - B^{2}) \\
\
\end{array}} \right\rbrack$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \theta = \cos^{- 1}\left\lbrack - \frac{A^{2} + B^{2}}{\begin{array}{r}
2(A^{2} - B^{2}) \\
\
\end{array}} \right\rbrack$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.