1. **Stating the problem:** We have three triangles with given angles and sides, and we want to analyze or solve for unknown sides or angles using trigonometry. 2. **Triangle Zone 5B:** Given angles $\angle D = 40^\circ$, $\angle F = 65^\circ$, and side $EF = 13.3$ hm. - First, find the third angle $\angle E$ using the triangle angle sum rule: $$\angle E = 180^\circ - 40^\circ - 65^\circ = 75^\circ$$ 3. Use the Law of Sines to find sides $DE$ and $DF$: $$\frac{DE}{\sin 65^\circ} = \frac{DF}{\sin 40^\circ} = \frac{13.3}{\sin 75^\circ}$$ Calculate each side: $$DE = \frac{13.3 \times \sin 65^\circ}{\sin 75^\circ}$$ $$DF = \frac{13.3 \times \sin 40^\circ}{\sin 75^\circ}$$ 4. **Triangle Zone 10A:** Right triangle with $\angle A = 55^\circ$ and side $AB = 10.8$ hm. - Since it's right angled, the other non-right angle is $35^\circ$. - Use trigonometric ratios to find other sides: $$BC = AB \times \tan 55^\circ$$ $$AC = AB / \cos 55^\circ$$ 5. **Triangle Zone 5C:** Given $\angle F = 62^\circ$ and side $FH = 19.5$ hm. - Without more information, we cannot solve further. **Summary:** - Zone 5B: $\angle E = 75^\circ$, $DE = \frac{13.3 \sin 65^\circ}{\sin 75^\circ}$ hm, $DF = \frac{13.3 \sin 40^\circ}{\sin 75^\circ}$ hm. - Zone 10A: $BC = 10.8 \times \tan 55^\circ$ hm, $AC = \frac{10.8}{\cos 55^\circ}$ hm. - Zone 5C: Insufficient data to solve. Final numeric approximations: - $DE \approx 12.7$ hm - $DF \approx 7.5$ hm - $BC \approx 15.3$ hm - $AC \approx 18.9$ hm