Resultant of two vectors of magnitudes P and Q is of magnitude ‘Q’. If the magnitude of $\overrightarrow{Q}$ is doubled now the angle made by new resultant with $\overrightarrow{P}$ is

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

Given $\text{R}=\text{Q}$

Now ${\text{R}}^{\text{2}}={\text{P}}^{\text{2}}+{\text{Q}}^{\text{2}}+2\text{PQ}\cos \theta$

$2\text{Q}\cos \theta =-\text{P}\to \left(1\right)$

Now $\overrightarrow{Q}$ is doubled the new resultant is ${\overrightarrow{R}}_1=\overrightarrow{P}+2\overrightarrow{Q}$

Then $\tan \alpha =\frac{2Q\cos \theta }{P+2Q\cos \theta }\to \left(2\right)$

From (1) & (2)

$\tan \alpha =\infty$

$\alpha =90°$