1. **Problem statement:** Two forces $P$ and $Q$ act on a particle at point $O$. Force $P$ has magnitude 15 N, force $Q$ has magnitude $X$ N, and the angle between $P$ and $Q$ is 150°. The resultant force $R$ makes an angle of 50° with $Q$. We need to find: (a) The magnitude of $R$. (b) The value of $X$. 2. **Formulas and rules:** - The magnitude of the resultant of two forces $P$ and $Q$ with angle $ heta$ between them is given by the law of cosines: $$R = \sqrt{P^2 + Q^2 + 2PQ\cos\theta}$$ - The angle $eta$ between $R$ and $Q$ satisfies the law of sines in the force triangle: $$\frac{P}{\sin\beta} = \frac{R}{\sin\theta} = \frac{Q}{\sin\alpha}$$ where $\alpha$ is the angle between $R$ and $P$. 3. **Given:** - $P = 15$ - $Q = X$ (unknown) - Angle between $P$ and $Q$, $\theta = 150^\circ$ - Angle between $R$ and $Q$, $\beta = 50^\circ$ 4. **Find $R$ in terms of $X$:** $$R = \sqrt{15^2 + X^2 + 2 \times 15 \times X \times \cos 150^\circ}$$ Since $\cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \approx -0.866$, $$R = \sqrt{225 + X^2 - 15X \times 1.732} = \sqrt{225 + X^2 - 25.98X}$$ 5. **Use the law of sines to relate $P$, $Q$, $R$, and angles:** The angle between $P$ and $R$ is $\alpha = 180^\circ - (150^\circ + 50^\circ) = -20^\circ$ but since angles in triangle sum to 180°, the angle opposite $Q$ is $\beta = 50^\circ$, opposite $P$ is $\alpha = 80^\circ$ (since $150^\circ$ is between $P$ and $Q$, the triangle formed by $P$, $Q$, and $R$ has angles $50^\circ$, $80^\circ$, and $50^\circ$). Actually, the angle opposite $P$ is $50^\circ$, opposite $Q$ is $80^\circ$, and opposite $R$ is $50^\circ$. But the problem states angle between $R$ and $Q$ is 50°, so angle opposite $P$ is 50°, angle opposite $Q$ is 80°, angle opposite $R$ is 50°. Using law of sines: $$\frac{P}{\sin 50^\circ} = \frac{Q}{\sin 80^\circ} = \frac{R}{\sin 50^\circ}$$ From this, since $P=15$: $$\frac{15}{\sin 50^\circ} = \frac{X}{\sin 80^\circ}$$ 6. **Calculate $X$:** $$X = 15 \times \frac{\sin 80^\circ}{\sin 50^\circ}$$ Using approximate values: $\sin 50^\circ \approx 0.7660$, $\sin 80^\circ \approx 0.9848$ $$X = 15 \times \frac{0.9848}{0.7660} = 15 \times 1.285 = 19.28$$ 7. **Calculate $R$ using $X=19.28$:** $$R = \sqrt{225 + (19.28)^2 - 25.98 \times 19.28}$$ Calculate each term: $19.28^2 = 371.7$ $25.98 \times 19.28 = 501.1$ So, $$R = \sqrt{225 + 371.7 - 501.1} = \sqrt{95.6} = 9.78$$ 8. **Final answers:** (a) Magnitude of $R$ is approximately $9.79$ N. (b) Magnitude of $Q$ is approximately $19.3$ N.