1. **Problem statement:** Given the coefficient of static friction $\mu = \sqrt{3}$ between a body and an inclined rough plane, find the angle of inclination $\theta$ at which the body is about to slide due to its weight alone. 2. **Understanding the physics:** The body is on the verge of sliding when the component of gravitational force down the slope equals the maximum static friction force. 3. **For an inclined plane:** - The component of weight down the slope is $mg \sin \theta$. - The normal force is $mg \cos \theta$. - The maximum static friction force is $f_s = \mu mg \cos \theta$. 4. **At the point of impending motion:** $$mg \sin \theta = \mu mg \cos \theta$$ 5. **Simplify by canceling $mg$:** $$\sin \theta = \mu \cos \theta$$ 6. **Divide both sides by $\cos \theta$:** $$\tan \theta = \mu$$ 7. **Substitute $\mu = \sqrt{3}$:** $$\tan \theta = \sqrt{3}$$ 8. **Find $\theta$:** $$\theta = \tan^{-1}(\sqrt{3}) = 60^\circ$$ **Final answer:** The inclination angle is $60^\circ$. Hence, the correct option is (c) 60°.