1. **State the problem:** We need to find the angle $\theta$ in the triangle formed by the sides 3 m, 4 m, and 5.5 m, where $\theta$ is the angle between the 4 m and 3 m sides. 2. **Identify the triangle type and sides:** The sides given are 3 m, 4 m, and 5.5 m. We want to find the angle opposite the 5.5 m side, which is $\theta$. 3. **Use the Law of Cosines:** The Law of Cosines relates the sides and angles of any triangle: $$c^2 = a^2 + b^2 - 2ab\cos(\theta)$$ where $c$ is the side opposite angle $\theta$, and $a$, $b$ are the other two sides. 4. **Plug in the values:** Here, $c = 5.5$, $a = 3$, and $b = 4$. $$5.5^2 = 3^2 + 4^2 - 2 \times 3 \times 4 \times \cos(\theta)$$ 5. **Calculate squares:** $$30.25 = 9 + 16 - 24 \cos(\theta)$$ 6. **Simplify the right side:** $$30.25 = 25 - 24 \cos(\theta)$$ 7. **Isolate the cosine term:** $$30.25 - 25 = -24 \cos(\theta)$$ $$5.25 = -24 \cos(\theta)$$ 8. **Divide both sides by -24:** $$\cancel{\frac{5.25}{-24}} = \cancel{\frac{-24 \cos(\theta)}{-24}}$$ $$\cos(\theta) = -\frac{5.25}{24} = -0.21875$$ 9. **Find the angle $\theta$ by taking the inverse cosine:** $$\theta = \cos^{-1}(-0.21875)$$ 10. **Calculate the angle:** $$\theta \approx 102.6^\circ$$ **Final answer:** The angle $\theta$ is approximately $102.6^\circ$.