The sum of two forces $\overrightarrow{P}$ and $\overrightarrow{Q}$ is $\overrightarrow{R}$ such that $\left|\overrightarrow{R}\right|=\left|\overrightarrow{P}\right|$. The angle $\alpha$ (in degrees) that the resultant of $2\overrightarrow{P}$ and $\overrightarrow{Q}$ will make with $\overrightarrow{Q}$ is ________

$\overrightarrow{R}=\overrightarrow{P}+\overrightarrow{Q}$

$P^2=P^2+Q^2+2PQ\cos \theta$

$\text{cos}\theta =\frac{-Q}{2P}\Rightarrow \sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{Q^2}{4P^2}}$

$\overrightarrow{R\text{'}}=\overrightarrow{2P}+\overrightarrow{Q}$    (Angle between $\overrightarrow{P}\&\overrightarrow{Q}$ is same as that between $2\overrightarrow{P}\&\overrightarrow{Q}$

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

$\tan \alpha =\frac{2P\sqrt{1-\frac{Q^2}{4P^2}}}{Q+2P\left(\frac{-Q}{2P}\right)}=\frac{\sqrt{4P^2-Q^2}}{0}=\infty$

$\Rightarrow \alpha =90°$