1. **State the problem:** A ring is held stationary by three strings with tensions $T_1$, $T_2$, and $T_3=10$ N. $T_1$ pulls due East, $T_2$ pulls at 30° West of North, and $T_3$ pulls due South. The system is in equilibrium, so the net force is zero. We need to find $T_1$ and $T_2$. 2. **Use equilibrium conditions:** For equilibrium, the sum of forces in both horizontal (x) and vertical (y) directions must be zero. 3. **Set up coordinate axes:** Let East be positive x-direction and North be positive y-direction. 4. **Resolve tensions into components:** - $T_1$ is due East: $T_1$ has components $(T_1, 0)$. - $T_2$ is 30° West of North: its components are $(-T_2 \sin 30^\circ, T_2 \cos 30^\circ)$. - $T_3$ is due South: components $(0, -10)$. 5. **Write equilibrium equations:** - Horizontal (x): $T_1 - T_2 \sin 30^\circ = 0$ - Vertical (y): $T_2 \cos 30^\circ - 10 = 0$ 6. **Calculate trigonometric values:** - $\sin 30^\circ = 0.5$ - $\cos 30^\circ = \frac{\sqrt{3}}{2} \approx 0.866$ 7. **Solve vertical equation for $T_2$:** $$ T_2 \times 0.866 = 10 \implies T_2 = \frac{10}{0.866} \approx 11.55 $$ 8. **Solve horizontal equation for $T_1$:** $$ T_1 = T_2 \times 0.5 = 11.55 \times 0.5 = 5.775 $$ 9. **Final answer:** - $T_1 \approx 5.78$ N - $T_2 \approx 11.55$ N