Two forces P and Q are in ratio P: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

$\frac{P}{Q}=\frac{1}{2},\alpha ={\tan }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$\frac{\sqrt{3}}{2}=\frac{Q\sin \theta }{P+Q\cos \theta }=\frac{\sin \theta }{\left(\frac{P}{Q}+\cos \theta \right)}=\frac{\sin \theta }{\left(\frac{1}{2}+\cos \theta \right)}$

$\frac{\sqrt{3}}{2}=\frac{2\sin \theta }{1+2\cos \theta }$

Eg on b.s

$\frac{3}{4}=\frac{4{\sin }^2\theta }{1+4{\cos }^2\theta +4\cos \theta }$

$\begin{array}{l}16-16{\cos }^2\theta =3+12{\cos }^2\theta +12\cos \theta \\ \Rightarrow 13=28{\cos }^2\theta +12\cos \theta \\ \therefore 28{\cos }^2\theta +12\cos \theta -13=0\end{array}$

$\begin{array}{l}\cos \theta =\frac{-12\pm \sqrt{144+4\left(28\right)\left(13\right)}}{2\times 28}\\ =\frac{-12\pm \sqrt{1600}}{2\times 28}=\frac{-12\pm 40}{56}\\ =\frac{28}{56}\\ \therefore \theta ={60}^0\end{array}$