Two forces P and Q are in ratio $\text{P}:\text{Q}=1:2$ . If their resultant is at an angle ${\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$ to vector P, then angle between P and Q is:

$\frac{\text{P}}{\text{Q}}=\frac{1}{2}$

$\text{Q}=2\text{P}$

$\theta ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\alpha ={\tan }^{-1}\left(\frac{\text{Q}\sin \theta }{\text{P}+\text{Q}\cos \theta }\right)$

$={\tan }^{-1}\left(\frac{2\text{P}\sin \theta }{\text{P}+2\text{P}\cos \theta }\right)$

$\tan \alpha =\frac{2\sin \theta }{1+2\cos \theta }$

$\frac{\sqrt{3}}{2}=\frac{2\sin \theta }{1+2\cos \theta }\left[\text{S}\text{.O}\text{.B}\text{.S}\right]$

$\frac{3}{4}=\frac{4{\sin }^2\theta }{{\left(1+2\cos \theta \right)}^2}$

$=\frac{4{\sin }^2\theta }{1+4{\cos }^2\theta +4\cos \theta }$

$\frac{3}{4}=\frac{4\left(1-{\cos }^2\theta \right)}{1+4{\cos }^2\theta +4\cos \theta }$

$3\left(1+4{\cos }^2\theta +4\cos \theta \right)=4.4\left(1-{\cos }^2\theta \right)$

$3+12{\cos }^2\theta +12\cos \theta =16-16{\cos }^2\theta$

$28{\cos }^2\theta +12\cos \theta -13=0$

$\text{W}\text{.K}\text{.T}$

$x=\frac{-\text{b}\pm \sqrt{{\text{b}}^2-4\text{ac}}}{2\text{a}}$

$\cos \theta =\frac{-12\pm \sqrt{{12}^2-4\times 28\left(-13\right)}}{2\times 28}$

$=\frac{-12\pm \sqrt{144+1450}}{56}$

$=\frac{-12\pm \sqrt{1600}}{56}=\frac{-12\pm 40}{56}$

$=\frac{-12+40}{56}=\frac{28}{56}=\frac{1}{2}$

$\cos \theta =\frac{1}{2}\Rightarrow \theta =60°$