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 $\theta$ (in degrees) that the resultants of $2\overrightarrow{P}$  and $\overrightarrow{Q}$  will make with  $\overrightarrow{Q}$ is ____

From triangle law of vector addition , we can say that angle between  $\left(2\overrightarrow{P}+\overrightarrow{Q}\right)$ and $\overrightarrow{Q}$  is 90º
 
 Alternate solution  $\left|\overrightarrow{P}+\overrightarrow{Q}\right|=\left|\overrightarrow{P}\right|$
              
 $$\begin{gathered} P^2+Q^2+2PQ\cos \theta =P^2 \\ \Rightarrow Q+2P\cos \theta =0 \\ \Rightarrow \cos \theta =\frac{-Q}{2P} \end{gathered}$$

 

$\tan \alpha =\frac{2P\sin \theta }{2P\cos \theta +Q}=\infty \because \left[2P\cos \theta +Q=0\right]$

$\alpha =90º$

So angle between  $2\overline{P}+\overline{Q}$ and  $\overline{Q}$ is 90º