Answer:
Clearly, \[{{P}^{2}}+{{\left( \frac{Q}{2} \right)}^{2}}={{Q}^{2}}\] or \[{{P}^{2}}=\frac{3}{4}{{Q}^{2}}\] or \[P=\frac{\sqrt{3}}{2}Q\] \[\therefore \] \[\tan \theta =\frac{Q/2}{P}=\frac{Q/2}{\sqrt{3}Q/2}=\frac{1}{\sqrt{3}}\] or \[\theta ={{30}^{\circ }}\] Angle between \[\vec{P}\]and \[\vec{Q},\] \[\beta ={{180}^{\circ }}-\theta ={{180}^{\circ }}-{{30}^{\circ }}=\mathbf{15}{{\mathbf{0}}^{\mathbf{o}}}\]
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