Q. The resultant of two forces at right angle is 5N. When the angle between them is 120${^\circ}$, the resultant is $\sqrt{13}$. Then the force are
A) $\sqrt{12}N,\sqrt{13}N$
B) $\sqrt{20}N,\sqrt{5}N$
C) 3 N, 4 N
D) $\sqrt{40}N,\sqrt{15}N$
Explanation
(c) Let $A$ and $B$ be the two forces. As per question
$\sqrt{A^{2} + B^{2}} = 5$
or $A^{2} + B^{2} = 25$ (i)
and $A^{2} + B^{2} + 2AB\cos 120{^\circ} = 13$
or $25 + 2AB \times ( - 1/2) = 13$
or $AB = 25 - 13 = 12$
or 2$AB = 24$ (ii)
Solving (i) and (ii), we get
$A = 3$N
and $B = 4$ N
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.