1. **Problem 1:** Given $A=40^\circ$, $B=80^\circ$, and $a=15$ cm, find $b$. 2. **Formula:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Step 1:** Find angle $C$ using the triangle angle sum rule: $$C = 180^\circ - A - B = 180^\circ - 40^\circ - 80^\circ = 60^\circ$$ 4. **Step 2:** Use the Law of Sines to find $b$: $$\frac{a}{\sin A} = \frac{b}{\sin B} \implies b = \frac{a \sin B}{\sin A}$$ 5. **Step 3:** Substitute values: $$b = \frac{15 \times \sin 80^\circ}{\sin 40^\circ}$$ 6. **Step 4:** Calculate sines: $$\sin 80^\circ \approx 0.9848, \quad \sin 40^\circ \approx 0.6428$$ 7. **Step 5:** Calculate $b$: $$b = \frac{15 \times 0.9848}{0.6428} = \frac{14.772}{0.6428} \approx 22.98 \text{ cm}$$ --- 8. **Problem 2:** Given $A=55^\circ$, $B=45^\circ$, and $a=12$ m, find $c$. 9. **Step 1:** Find angle $C$: $$C = 180^\circ - 55^\circ - 45^\circ = 80^\circ$$ 10. **Step 2:** Use Law of Sines to find $c$: $$\frac{a}{\sin A} = \frac{c}{\sin C} \implies c = \frac{a \sin C}{\sin A}$$ 11. **Step 3:** Substitute values: $$c = \frac{12 \times \sin 80^\circ}{\sin 55^\circ}$$ 12. **Step 4:** Calculate sines: $$\sin 80^\circ \approx 0.9848, \quad \sin 55^\circ \approx 0.8192$$ 13. **Step 5:** Calculate $c$: $$c = \frac{12 \times 0.9848}{0.8192} = \frac{11.8176}{0.8192} \approx 14.43 \text{ m}$$ --- 14. **Problem 3:** Given $A=30^\circ$, $a=8$, and $b=10$, find $B$. 15. **Step 1:** Use Law of Sines: $$\frac{a}{\sin A} = \frac{b}{\sin B} \implies \sin B = \frac{b \sin A}{a}$$ 16. **Step 2:** Substitute values: $$\sin B = \frac{10 \times \sin 30^\circ}{8}$$ 17. **Step 3:** Calculate $\sin 30^\circ$: $$\sin 30^\circ = 0.5$$ 18. **Step 4:** Calculate $\sin B$: $$\sin B = \frac{10 \times 0.5}{8} = \frac{5}{8} = 0.625$$ 19. **Step 5:** Find $B$ by taking inverse sine: $$B = \sin^{-1}(0.625) \approx 38.68^\circ$$ 20. **Step 6:** Check for possible ambiguous case (since sine is positive in first and second quadrants): $$B' = 180^\circ - 38.68^\circ = 141.32^\circ$$ 21. **Step 7:** Check if $A + B' < 180^\circ$: $$30^\circ + 141.32^\circ = 171.32^\circ < 180^\circ$$ So both $B = 38.68^\circ$ and $B = 141.32^\circ$ are possible solutions depending on the triangle. **Final answers:** 1. $b \approx 22.98$ cm 2. $c \approx 14.43$ m 3. $B \approx 38.68^\circ$ or $141.32^\circ$