1. **Problem statement:** A body of weight $W$ is on a rough horizontal plane with coefficient of static friction $\mu = \frac{\sqrt{3}}{3}$. A horizontal force $F$ acts on the body, and the body is in equilibrium. We need to find the angle $\theta$ between the resultant reaction $R$ and the normal reaction $N$. 2. **Forces involved:** - Weight $W$ acts vertically downward. - Normal reaction $N$ acts vertically upward. - Frictional force $f$ acts horizontally opposite to $F$. - Resultant reaction $R$ is the vector sum of $N$ and $f$. 3. **Equilibrium conditions:** Since the body is in equilibrium, the frictional force $f$ balances the horizontal force $F$, so $f = F$. 4. **Maximum static friction:** The maximum friction force is $f_{max} = \mu N$. 5. **Relation between forces:** Since the body is in equilibrium and friction is static, $f \leq \mu N$. 6. **Angle between resultant reaction and normal reaction:** The resultant reaction $R$ is the vector sum of $N$ (vertical) and $f$ (horizontal). The angle $\theta$ between $R$ and $N$ satisfies: $$\tan \theta = \frac{f}{N}$$ 7. **Using maximum friction:** At equilibrium with impending motion, $f = \mu N$, so: $$\tan \theta = \mu = \frac{\sqrt{3}}{3}$$ 8. **Calculate $\theta$:** $$\theta = \arctan\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$$ 9. **Interpretation:** The angle $\theta$ is exactly $\frac{\pi}{6}$, so the measure of the angle between the resultant reaction and the normal reaction lies in the interval $]0, \frac{\pi}{6}]$. **Final answer:** (c) $]0, \pi/6]$