1. **Problem:** In triangle ABC, given angles $A=42^\circ$, $B=58^\circ$, and side $a=15$ cm, find sides $b$ and $c$. 2. **Formula:** Use the Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Step 1:** Find angle $C$ using the triangle angle sum rule: $$C = 180^\circ - A - B = 180^\circ - 42^\circ - 58^\circ = 80^\circ$$ 4. **Step 2:** Calculate side $b$: $$b = \frac{a \sin B}{\sin A} = \frac{15 \sin 58^\circ}{\sin 42^\circ}$$ 5. **Step 3:** Calculate side $c$: $$c = \frac{a \sin C}{\sin A} = \frac{15 \sin 80^\circ}{\sin 42^\circ}$$ 6. **Step 4:** Evaluate values: $$\sin 42^\circ \approx 0.6691, \sin 58^\circ \approx 0.8480, \sin 80^\circ \approx 0.9848$$ 7. **Step 5:** Substitute: $$b = \frac{15 \times 0.8480}{0.6691} \approx 19.01 \text{ cm}$$ $$c = \frac{15 \times 0.9848}{0.6691} \approx 22.07 \text{ cm}$$ --- 8. **Problem:** Two sides of a triangle measure 18 m and 25 m with included angle $65^\circ$. Find the third side. 9. **Formula:** Use the Law of Cosines: $$c^2 = a^2 + b^2 - 2ab \cos C$$ 10. **Step 1:** Assign $a=18$, $b=25$, $C=65^\circ$. 11. **Step 2:** Calculate $c^2$: $$c^2 = 18^2 + 25^2 - 2 \times 18 \times 25 \times \cos 65^\circ$$ 12. **Step 3:** Evaluate: $$18^2 = 324, \quad 25^2 = 625, \quad \cos 65^\circ \approx 0.4226$$ 13. **Step 4:** Substitute: $$c^2 = 324 + 625 - 2 \times 18 \times 25 \times 0.4226 = 949 - 380.34 = 568.66$$ 14. **Step 5:** Find $c$: $$c = \sqrt{568.66} \approx 23.85 \text{ m}$$ --- 15. **Problem:** The sides of a triangle are 9 cm, 12 cm, and 16 cm. Find the largest angle. 16. **Formula:** Use Law of Cosines to find angle opposite largest side $c=16$: $$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$$ 17. **Step 1:** Assign $a=9$, $b=12$, $c=16$. 18. **Step 2:** Calculate: $$\cos C = \frac{9^2 + 12^2 - 16^2}{2 \times 9 \times 12} = \frac{81 + 144 - 256}{216} = \frac{-31}{216} \approx -0.1435$$ 19. **Step 3:** Find $C$: $$C = \cos^{-1}(-0.1435) \approx 98.25^\circ$$ --- 20. **Problem:** A plane flies 120 km on a bearing of $040^\circ$ and then 150 km on a bearing of $110^\circ$. Find the distance from the starting point. 21. **Step 1:** Find the angle between the two legs: $$\theta = 110^\circ - 40^\circ = 70^\circ$$ 22. **Step 2:** Use Law of Cosines for distance $d$: $$d^2 = 120^2 + 150^2 - 2 \times 120 \times 150 \times \cos 70^\circ$$ 23. **Step 3:** Calculate: $$120^2 = 14400, \quad 150^2 = 22500, \quad \cos 70^\circ \approx 0.3420$$ 24. **Step 4:** Substitute: $$d^2 = 14400 + 22500 - 2 \times 120 \times 150 \times 0.3420 = 36900 - 12312 = 24588$$ 25. **Step 5:** Find $d$: $$d = \sqrt{24588} \approx 156.8 \text{ km}$$ --- 26. **Problem:** Given $a=12$, $b=20$, and $A=35^\circ$, determine all possible triangles. 27. **Step 1:** Use Law of Sines to find $\sin B$: $$\frac{a}{\sin A} = \frac{b}{\sin B} \Rightarrow \sin B = \frac{b \sin A}{a} = \frac{20 \sin 35^\circ}{12}$$ 28. **Step 2:** Calculate: $$\sin 35^\circ \approx 0.574, \quad \sin B = \frac{20 \times 0.574}{12} = 0.957$$ 29. **Step 3:** Since $\sin B = 0.957$, possible $B$ values are: $$B_1 = \sin^{-1}(0.957) \approx 73.74^\circ$$ $$B_2 = 180^\circ - 73.74^\circ = 106.26^\circ$$ 30. **Step 4:** For each $B$, find $C$: $$C_1 = 180^\circ - 35^\circ - 73.74^\circ = 71.26^\circ$$ $$C_2 = 180^\circ - 35^\circ - 106.26^\circ = 38.74^\circ$$ 31. **Step 5:** Find side $c$ for each triangle using Law of Sines: $$c_1 = \frac{a \sin C_1}{\sin A} = \frac{12 \sin 71.26^\circ}{\sin 35^\circ}$$ $$c_2 = \frac{12 \sin 38.74^\circ}{\sin 35^\circ}$$ 32. **Step 6:** Calculate: $$\sin 71.26^\circ \approx 0.949, \sin 38.74^\circ \approx 0.626$$ 33. **Step 7:** Substitute: $$c_1 = \frac{12 \times 0.949}{0.574} \approx 19.84$$ $$c_2 = \frac{12 \times 0.626}{0.574} \approx 13.09$$ **Final answers:** 1) $b \approx 19.01$ cm, $c \approx 22.07$ cm 2) Third side $\approx 23.85$ m 3) Largest angle $\approx 98.25^\circ$ 4) Distance from start $\approx 156.8$ km 5) Two possible triangles: - Triangle 1: $B \approx 73.74^\circ$, $C \approx 71.26^\circ$, $c \approx 19.84$ - Triangle 2: $B \approx 106.26^\circ$, $C \approx 38.74^\circ$, $c \approx 13.09$