Question

What is the angle between two forces of equal magnitude such that the resultant is one-third as much as either of the original forces?

• A. ${\cos }^{-1}(-\frac{17}{18})$
• B. ${\cos }^{-1}\left(\frac{1}{3}\right)$
• C. ${45}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is A ${\cos }^{-1}(-\frac{17}{18})$

Let the vectors be $\overrightarrow{A}$ & $\overrightarrow{B}$. Then,
$R=\sqrt{A^2+B^2+2AB\cos \theta }$
$R^2=A^2+B^2+2AB\cos \theta$

Given that $A=B$ & $R=\frac{A}{3}$
Putting the values of $A,B$ and $R$,
${\left(\frac{A}{3}\right)}^2=A^2+A^2+2\times A\times A\cos \theta$
$\Rightarrow \frac{1}{9}=2(1+\cos \theta )$
$\Rightarrow 1+\cos \theta =\frac{1}{18}$
$\Rightarrow \cos \theta =\frac{1}{18}-1=-\frac{17}{18}$
$\Rightarrow \theta ={\cos }^{-1}(-\frac{17}{18})$