1. **Problem 6:** Find the distance between towns B and C in triangle ABC where AB = 45 km, $\angle A = 37^\circ$, and $\angle C = 110^\circ$. 2. Use the Law of Sines formula: $$\frac{AB}{\sin \angle C} = \frac{BC}{\sin \angle A}$$ 3. Substitute known values: $$\frac{45}{\sin 110^\circ} = \frac{BC}{\sin 37^\circ}$$ 4. Solve for BC: $$BC = \frac{45 \times \sin 37^\circ}{\sin 110^\circ}$$ 5. Calculate sine values: $$\sin 37^\circ \approx 0.6018, \quad \sin 110^\circ \approx 0.9397$$ 6. Compute BC: $$BC = \frac{45 \times 0.6018}{0.9397} = \frac{27.081}{0.9397} \approx 28.81 \text{ km}$$ --- 1. **Problem 7:** Given triangle JKL with sides JK = 3 cm, KL = 7 cm, and $\angle J = 101^\circ$, find side k (JL), $\angle L$, and $\angle J$. 2. Use the Law of Cosines to find side k: $$k^2 = JK^2 + KL^2 - 2 \times JK \times KL \times \cos \angle J$$ 3. Substitute values: $$k^2 = 3^2 + 7^2 - 2 \times 3 \times 7 \times \cos 101^\circ$$ 4. Calculate: $$k^2 = 9 + 49 - 42 \times \cos 101^\circ$$ 5. Since $\cos 101^\circ \approx -0.1908$, $$k^2 = 58 - 42 \times (-0.1908) = 58 + 8.0136 = 66.0136$$ 6. Find k: $$k = \sqrt{66.0136} \approx 8.12 \text{ cm}$$ 7. Use Law of Sines to find $\angle L$: $$\frac{k}{\sin 101^\circ} = \frac{JK}{\sin L}$$ 8. Substitute known values: $$\frac{8.12}{\sin 101^\circ} = \frac{3}{\sin L}$$ 9. Calculate $\sin 101^\circ \approx 0.9816$, so: $$\frac{8.12}{0.9816} = \frac{3}{\sin L} \Rightarrow 8.27 = \frac{3}{\sin L}$$ 10. Solve for $\sin L$: $$\sin L = \frac{3}{8.27} \approx 0.3626$$ 11. Find $\angle L$: $$L = \sin^{-1}(0.3626) \approx 21.3^\circ$$ 12. Find $\angle K$ using triangle angle sum: $$\angle K = 180^\circ - 101^\circ - 21.3^\circ = 57.7^\circ$$ --- 1. **Problem 8:** Two friends bike from the same point. Alex bikes 3.8 km north, Jordan bikes 5.1 km in another direction, and the distance between them is 6.2 km. Find the angle between their routes. 2. Use Law of Cosines for triangle with sides 3.8 km, 5.1 km, and 6.2 km: $$c^2 = a^2 + b^2 - 2ab \cos \theta$$ 3. Let $\theta$ be the angle between the two routes, with sides $a=3.8$, $b=5.1$, and opposite side $c=6.2$. 4. Substitute values: $$6.2^2 = 3.8^2 + 5.1^2 - 2 \times 3.8 \times 5.1 \times \cos \theta$$ 5. Calculate squares: $$38.44 = 14.44 + 26.01 - 38.76 \cos \theta$$ 6. Simplify: $$38.44 = 40.45 - 38.76 \cos \theta$$ 7. Rearrange: $$38.76 \cos \theta = 40.45 - 38.44 = 2.01$$ 8. Solve for $\cos \theta$: $$\cos \theta = \frac{2.01}{38.76} \approx 0.0519$$ 9. Find $\theta$: $$\theta = \cos^{-1}(0.0519) \approx 87.0^\circ$$ --- **Final answers:** - Distance between towns B and C: $28.81$ km - Side k (JL): $8.12$ cm - $\angle L = 21.3^\circ$ - $\angle K = 57.7^\circ$ - Angle between friends' routes: $87.0^\circ$