Two Forces P and Q of magnitude 4F and 6F respectively are at angle $\theta$ with each other. If the force Q is double then their resultant also gets double then the angle $\theta$ is

Resultant of two force is given $R=\sqrt{P^2+Q^2+2PQ\cos \theta }$ 
$R^1=\sqrt{P^2+{\left(2Q\right)}^2+2P\left(2Q\right)\cos \theta }$ 
$\cos 4R^2=R^1=\sqrt{P^2+4Q^2+4PQ\cos \theta }=\frac{n^2-1}{n^2+1}$ 
$4(P^2+Q^2+2PQ\cos \theta )=P^2+4Q^2+4PQ\cos \theta$ 
$4P^2+8PQ\cos \theta =P^2+4PQ\cos \theta$ 
$\begin{array}{l}3P^2+4P\theta \cos \theta =0\\ 4P\theta \cos \theta =3P^2\end{array}$ 
$\cos \theta =-\frac{3P}{4\theta }$ 
$\cos \theta =-\frac{3\left(4F\right)}{4\left(6F\right)}$ 
$\cos \theta =-\frac{12}{24}=-\frac{1}{2}$