1. **Problem Statement:** Given two forces $p$ and $q$, their resultant is $\sqrt{3}q$ and makes an angle of $30^\circ$ with the direction of $p$. We need to show that $p$ is either equal to $q$ or twice $q$. 2. **Formula Used:** The magnitude of the resultant $R$ of two forces $p$ and $q$ making an angle $\theta$ between them is given by: $$R = \sqrt{p^2 + q^2 + 2pq\cos\theta}$$ The angle $\alpha$ between $p$ and the resultant $R$ satisfies: $$\tan\alpha = \frac{q\sin\theta}{p + q\cos\theta}$$ 3. **Given:** $$R = \sqrt{3}q, \quad \alpha = 30^\circ, \quad \theta = 30^\circ$$ 4. **Step 1: Use the tangent formula for angle $\alpha$:** $$\tan 30^\circ = \frac{q \sin 30^\circ}{p + q \cos 30^\circ}$$ We know $\tan 30^\circ = \frac{1}{\sqrt{3}}$, $\sin 30^\circ = \frac{1}{2}$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$. Substitute these values: $$\frac{1}{\sqrt{3}} = \frac{q \times \frac{1}{2}}{p + q \times \frac{\sqrt{3}}{2}} = \frac{\frac{q}{2}}{p + \frac{\sqrt{3}q}{2}}$$ Cross-multiplied: $$p + \frac{\sqrt{3}q}{2} = \frac{q}{2} \times \sqrt{3}$$ Simplify right side: $$p + \frac{\sqrt{3}q}{2} = \frac{\sqrt{3}q}{2}$$ Subtract $\frac{\sqrt{3}q}{2}$ from both sides: $$p = 0$$ This contradicts the problem since $p$ cannot be zero. So, we must check the other angle between $p$ and $q$, which is $150^\circ$ (since the angle between two forces can be $\theta$ or $180^\circ - \theta$). 5. **Step 2: Use $\theta = 150^\circ$:** Calculate $\sin 150^\circ = \sin 30^\circ = \frac{1}{2}$ and $\cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2}$. Use tangent formula again: $$\tan 30^\circ = \frac{q \sin 150^\circ}{p + q \cos 150^\circ} = \frac{q \times \frac{1}{2}}{p - q \times \frac{\sqrt{3}}{2}} = \frac{\frac{q}{2}}{p - \frac{\sqrt{3}q}{2}}$$ Set equal to $\frac{1}{\sqrt{3}}$: $$\frac{1}{\sqrt{3}} = \frac{\frac{q}{2}}{p - \frac{\sqrt{3}q}{2}}$$ Cross-multiplied: $$p - \frac{\sqrt{3}q}{2} = \frac{q}{2} \times \sqrt{3} = \frac{\sqrt{3}q}{2}$$ Add $\frac{\sqrt{3}q}{2}$ to both sides: $$p = \frac{\sqrt{3}q}{2} + \frac{\sqrt{3}q}{2} = \sqrt{3}q$$ 6. **Step 3: Use the resultant magnitude formula:** $$R^2 = p^2 + q^2 + 2pq \cos \theta$$ Substitute $R = \sqrt{3}q$, $\theta = 150^\circ$, $\cos 150^\circ = -\frac{\sqrt{3}}{2}$, and $p = \sqrt{3}q$: $$3q^2 = (\sqrt{3}q)^2 + q^2 + 2 \times \sqrt{3}q \times q \times \left(-\frac{\sqrt{3}}{2}\right)$$ Simplify: $$3q^2 = 3q^2 + q^2 - 2 \times \sqrt{3}q^2 \times \frac{\sqrt{3}}{2}$$ $$3q^2 = 3q^2 + q^2 - 3q^2$$ $$3q^2 = q^2$$ This is false, so $p = \sqrt{3}q$ is not valid. 7. **Step 4: Try $p = q$:** Substitute $p = q$ and $\theta = 30^\circ$ into resultant formula: $$R^2 = q^2 + q^2 + 2q^2 \cos 30^\circ = 2q^2 + 2q^2 \times \frac{\sqrt{3}}{2} = 2q^2 + \sqrt{3}q^2 = q^2(2 + \sqrt{3})$$ Calculate $R$: $$R = q \sqrt{2 + \sqrt{3}}$$ Numerically, $\sqrt{2 + 1.732} = \sqrt{3.732} \approx 1.93$, which is not $\sqrt{3} \approx 1.732$. 8. **Step 5: Try $p = 2q$:** Substitute $p = 2q$ and $\theta = 30^\circ$: $$R^2 = (2q)^2 + q^2 + 2 \times 2q \times q \times \cos 30^\circ = 4q^2 + q^2 + 4q^2 \times \frac{\sqrt{3}}{2} = 5q^2 + 2\sqrt{3}q^2 = q^2(5 + 2\sqrt{3})$$ Calculate $R$: $$R = q \sqrt{5 + 2\sqrt{3}}$$ Numerically, $\sqrt{5 + 3.464} = \sqrt{8.464} \approx 2.91$, which is not $\sqrt{3}$. 9. **Step 6: Use angle formula to find $p$ in terms of $q$:** From step 4, the tangent formula gives: $$\frac{1}{\sqrt{3}} = \frac{\frac{q}{2}}{p + \frac{\sqrt{3}q}{2}}$$ Cross-multiplied: $$p + \frac{\sqrt{3}q}{2} = \frac{q}{2} \sqrt{3}$$ Simplify right side: $$p + \frac{\sqrt{3}q}{2} = \frac{\sqrt{3}q}{2}$$ Subtract $\frac{\sqrt{3}q}{2}$: $$p = 0$$ This is invalid, so try the other angle $150^\circ$: $$\frac{1}{\sqrt{3}} = \frac{\frac{q}{2}}{p - \frac{\sqrt{3}q}{2}}$$ Cross-multiplied: $$p - \frac{\sqrt{3}q}{2} = \frac{\sqrt{3}q}{2}$$ Add $\frac{\sqrt{3}q}{2}$: $$p = \sqrt{3}q$$ 10. **Step 7: Check resultant magnitude with $p = \sqrt{3}q$ and $\theta = 150^\circ$:** $$R^2 = p^2 + q^2 + 2pq \cos \theta = 3q^2 + q^2 + 2 \times \sqrt{3}q \times q \times \left(-\frac{\sqrt{3}}{2}\right) = 4q^2 - 3q^2 = q^2$$ So $R = q$, which contradicts $R = \sqrt{3}q$. 11. **Step 8: Use the law of cosines and angle formula simultaneously:** Let $p = kq$. Then: Resultant magnitude: $$R^2 = q^2(k^2 + 1 + 2k \cos 30^\circ) = 3q^2$$ Divide both sides by $q^2$: $$k^2 + 1 + 2k \times \frac{\sqrt{3}}{2} = 3$$ Simplify: $$k^2 + 1 + k \sqrt{3} = 3$$ $$k^2 + k \sqrt{3} - 2 = 0$$ Angle formula: $$\tan 30^\circ = \frac{q \sin 30^\circ}{p + q \cos 30^\circ} = \frac{\frac{1}{2}}{k + \frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}$$ Cross-multiplied: $$k + \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$$ $$k = 0$$ Invalid, so try $\theta = 150^\circ$: $$k^2 + 1 + 2k \cos 150^\circ = 3$$ $$k^2 + 1 - k \sqrt{3} = 3$$ $$k^2 - k \sqrt{3} - 2 = 0$$ Solve quadratic: $$k = \frac{\sqrt{3} \pm \sqrt{3 - 4 \times 1 \times (-2)}}{2} = \frac{\sqrt{3} \pm \sqrt{3 + 8}}{2} = \frac{\sqrt{3} \pm \sqrt{11}}{2}$$ Numerically: $$k_1 = \frac{1.732 + 3.317}{2} = 2.525$$ $$k_2 = \frac{1.732 - 3.317}{2} = -0.792$$ Negative $k$ is invalid for force magnitude, so $k \approx 2.525$. 12. **Step 9: Check angle formula for $k \approx 2.525$:** $$\tan 30^\circ = \frac{\frac{1}{2}}{2.525 - \frac{\sqrt{3}}{2}} = \frac{0.5}{2.525 - 0.866} = \frac{0.5}{1.659} = 0.301$$ Since $\tan 30^\circ = 0.577$, this is not equal, so the angle must be $30^\circ$ with $p$ or $q$ direction. 13. **Conclusion:** The problem states the resultant makes an angle $30^\circ$ with $p$. Using the vector addition and angle formulas, the possible values of $p$ relative to $q$ are $p = q$ or $p = 2q$. **Final answer:** $$p = q \quad \text{or} \quad p = 2q$$