1. **Problem Statement:** Determine whether to use the Law of Sines or Law of Cosines to find the indicated side length or angle measure in each triangle. 2. **Formulas:** - Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ used when you know either two angles and one side (AAS or ASA) or two sides and a non-included angle (SSA). - Law of Cosines: $$c^2 = a^2 + b^2 - 2ab\cos C$$ used when you know two sides and the included angle (SAS) or all three sides (SSS). 3. **Part a) Find \(\angle F\) in triangle DEF:** - Given sides: DE = 24 m, DF = 20 m, EF = 35 m. - All three sides are known, no angles except \(\angle F\) which we want. - Use Law of Cosines to find \(\angle F\) because we have SSS. 4. **Calculate \(\angle F\):** $$\cos F = \frac{DE^2 + DF^2 - EF^2}{2 \times DE \times DF} = \frac{24^2 + 20^2 - 35^2}{2 \times 24 \times 20} = \frac{576 + 400 - 1225}{960} = \frac{-249}{960} = -0.259375$$ $$\angle F = \cos^{-1}(-0.259375) \approx 105.0^\circ$$ 5. **Part b) Find side \(p = PQ\) in triangle PQR:** - Given: PR = 30 km, \(\angle R = 62^\circ\), \(\angle Q = 35^\circ\), side \(PQ = p\) unknown. - Two angles and one side known (AAS), use Law of Sines. 6. **Calculate \(p\):** - Find \(\angle P = 180^\circ - 62^\circ - 35^\circ = 83^\circ\). - Using Law of Sines: $$\frac{p}{\sin 62^\circ} = \frac{30}{\sin 83^\circ} \Rightarrow p = \frac{30 \times \sin 62^\circ}{\sin 83^\circ}$$ - Calculate: $$p \approx \frac{30 \times 0.8829}{0.9925} \approx 26.7 \text{ km}$$ 7. **Part c) Find \(\angle B\) in triangle ABC:** - Given: AC = 20 cm, CB = 22 cm, \(\angle C = 95^\circ\), \(\angle B\) unknown. - Two sides and included angle known (SAS), use Law of Cosines. 8. **Calculate \(\angle B\):** - Use Law of Cosines on side AB (opposite \(\angle C\)) to find side AB first: $$AB^2 = AC^2 + CB^2 - 2 \times AC \times CB \times \cos 95^\circ$$ $$AB^2 = 20^2 + 22^2 - 2 \times 20 \times 22 \times \cos 95^\circ = 400 + 484 - 880 \times (-0.0872) = 884 + 76.74 = 960.74$$ $$AB = \sqrt{960.74} \approx 31.0 \text{ cm}$$ - Now use Law of Cosines to find \(\angle B\): $$\cos B = \frac{AC^2 + AB^2 - CB^2}{2 \times AC \times AB} = \frac{20^2 + 31.0^2 - 22^2}{2 \times 20 \times 31.0} = \frac{400 + 961 - 484}{1240} = \frac{877}{1240} = 0.7073$$ $$\angle B = \cos^{-1}(0.7073) \approx 45.0^\circ$$ **Final answers:** - a) Use Law of Cosines; \(\angle F \approx 105^\circ\) - b) Use Law of Sines; \(p \approx 26.7\) km - c) Use Law of Cosines; \(\angle B \approx 45^\circ\)