Question

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

• A. ${\cos }^{-1}(-\frac{17}{18})$
• B. ${\cos }^{-1}\left(\frac{-31}{32}\right)$
• C. ${\cos }^{-1}\left(\frac{15}{32}\right)$
• D. ${\cos }^{-1}\left(\frac{29}{32}\right)$

Answer 1

The correct option is B ${\cos }^{-1}\left(\frac{-31}{32}\right)$

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 $A=B$ & $R=\frac{A}{4}$
Putting the values of $A,B$ and $R$,
${\left(\frac{A}{4}\right)}^2=A^2+A^2+2\times A\times A\cos \theta$
$\Rightarrow \frac{1}{16}=2(1+\cos \theta )$
$\Rightarrow 1+\cos \theta =\frac{1}{32}$
$\Rightarrow \cos \theta =\frac{1}{32}-1=-\frac{31}{32}$
$\Rightarrow \theta ={\cos }^{-1}(-\frac{31}{32})$

Hence option B is the correct answer