If$\overrightarrow{P}-\overrightarrow{Q}=\overrightarrow{R}\text{and}\text{P}=\text{Q}=\text{R}$ , the ratio of angles between $\overrightarrow{P}\&\overrightarrow{R}\text{to}\overrightarrow{P}\&\overrightarrow{Q}$ is

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

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

$\left|\overrightarrow{P}-\overrightarrow{R}\right|=\overrightarrow{Q}$

$\sqrt{{\text{P}}^2+{\text{R}}^{\text{2}}-2\text{PR}\cos \theta }=\text{Q}$

$\text{=}\sqrt{2{\text{Q}}^{\text{2}}-2{\text{Q}}^{\text{2}}\cos \theta }=\text{Q}$

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

$2\left(1-\cos \theta \right)=1$

$2\left(2{\sin }^2\frac{\theta }{2}\right)=1$

$2{\sin }^2\frac{\theta }{2}=\frac{1}{2}$

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

$\sin \frac{\theta }{2}=\frac{1}{2}$

$\sin \frac{\theta }{2}=\sin 30°$

$\frac{\theta }{2}=30°$

$\theta =60°$

$\left|\overline{P}-\overline{Q}\right|={\text{R}}^{\text{2}}$

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

$\text{=}\sqrt{2{\text{Q}}^{\text{2}}-2{\text{Q}}^{\text{2}}\cos \theta }=\text{Q}$

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

$2\left(1-\cos \theta \right)=1$

$1-\cos \theta =\frac{1}{2}$

$2{\sin }^2\frac{\theta }{2}=\frac{1}{2}$

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

$\sin \frac{\theta }{2}=\frac{1}{2}$

$\sin \frac{\theta }{2}=\sin 30°$

$\frac{\theta }{2}=30$

$\theta =60°$

The ratio of angles between $\overrightarrow{P}\&\overrightarrow{R}\text{to}\overrightarrow{P}\&\overrightarrow{Q}$ is

$\overrightarrow{P}\&\overrightarrow{R}=60°$

$\overrightarrow{P}\&\overrightarrow{Q}=60°$

$\overrightarrow{P}\&\overrightarrow{R}:\overrightarrow{P}\&\overrightarrow{Q}=60°:60°$

$\overrightarrow{P}\&\overrightarrow{R}:\overrightarrow{P}\&\overrightarrow{Q}=1:1$