1. **Stating the problem:** We are given a system of equations involving forces $FA$, $FB$, and an angle $\theta$: $$ 8\sqrt{6} + FB \sqrt{3} + FB \cos \theta = 10 $$ $$ (\sqrt{3} + \cos \theta) FB = 10 $$ $$ FA = 2FB = 20 $$ We want to analyze and verify these relationships and solve for $FA$, $FB$, and $\theta$. 2. **Using the given relations:** From $FA = 2FB = 20$, we get: $$ FB = \frac{FA}{2} = 10 $$ 3. **From the second equation:** $$ (\sqrt{3} + \cos \theta) FB = 10 $$ Substitute $FB = 10$: $$ (\sqrt{3} + \cos \theta) \times 10 = 10 $$ Divide both sides by 10: $$ \cancel{10} (\sqrt{3} + \cos \theta) = \cancel{10} $$ $$ \sqrt{3} + \cos \theta = 1 $$ 4. **Solve for $\cos \theta$:** $$ \cos \theta = 1 - \sqrt{3} $$ 5. **From the first equation:** $$ 8\sqrt{6} + FB \sqrt{3} + FB \cos \theta = 10 $$ Substitute $FB = 10$ and $\cos \theta = 1 - \sqrt{3}$: $$ 8\sqrt{6} + 10 \sqrt{3} + 10 (1 - \sqrt{3}) = 10 $$ Simplify: $$ 8\sqrt{6} + 10 \sqrt{3} + 10 - 10 \sqrt{3} = 10 $$ $$ 8\sqrt{6} + 10 = 10 $$ Subtract 10 from both sides: $$ 8\sqrt{6} = 0 $$ This is false, so the initial equation is inconsistent with the others. 6. **Check the magnitude relation:** Given: $$ 10 = \sqrt{FA^2 + FB^2 + 2 FA FB \cos 30^\circ} $$ Substitute $FA=20$, $FB=10$, and $\cos 30^\circ = \frac{\sqrt{3}}{2}$: $$ 10 = \sqrt{20^2 + 10^2 + 2 \times 20 \times 10 \times \frac{\sqrt{3}}{2}} $$ Calculate inside the root: $$ = \sqrt{400 + 100 + 200 \sqrt{3}} $$ $$ = \sqrt{500 + 200 \sqrt{3}} $$ Numerically, $200 \sqrt{3} \approx 346.41$, so: $$ \sqrt{500 + 346.41} = \sqrt{846.41} \approx 29.09 $$ This does not equal 10, so the magnitude relation is also inconsistent. **Final conclusion:** The given equations are inconsistent with each other under the assumptions $FA=20$ and $FB=10$. The only consistent relation found is: $$ \cos \theta = 1 - \sqrt{3} $$ which is approximately $-0.732$. **Answer:** $$\cos \theta = 1 - \sqrt{3}$$