1. **State the problem:** Given a triangle with side $a=18$ cm, side $c=18.9$ cm, and angle $A=66^\circ$, find all possible triangles (i.e., find angle $C$, angle $B$, and side $b$). 2. **Identify the knowns and unknowns:** We know $a$, $c$, and angle $A$. We want to find angles $B$, $C$ and side $b$. 3. **Use the Law of Sines:** The Law of Sines states $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}.$$ 4. **Find angle $C$ using Law of Sines:** $$\frac{a}{\sin A} = \frac{c}{\sin C} \implies \sin C = \frac{c \sin A}{a} = \frac{18.9 \times \sin 66^\circ}{18}.$$ Calculate $\sin 66^\circ \approx 0.9135$, so $$\sin C = \frac{18.9 \times 0.9135}{18} = \frac{17.27}{18} = 0.9594.$$ 5. **Check for possible triangles:** Since $\sin C = 0.9594$, angle $C$ can be $$C_1 = \arcsin(0.9594) \approx 73.8^\circ,$$ or the supplementary angle $$C_2 = 180^\circ - 73.8^\circ = 106.2^\circ.$$ 6. **Find angle $B$ for each case:** Since the sum of angles in a triangle is $180^\circ$, - For $C_1 = 73.8^\circ$, $$B_1 = 180^\circ - 66^\circ - 73.8^\circ = 40.2^\circ.$$ - For $C_2 = 106.2^\circ$, $$B_2 = 180^\circ - 66^\circ - 106.2^\circ = 7.8^\circ.$$ 7. **Find side $b$ for each case using Law of Sines:** $$b = \frac{a \sin B}{\sin A}.$$ - For $B_1 = 40.2^\circ$, $$b_1 = \frac{18 \times \sin 40.2^\circ}{\sin 66^\circ} = \frac{18 \times 0.646}{0.9135} = 12.7 \text{ cm}.$$ - For $B_2 = 7.8^\circ$, $$b_2 = \frac{18 \times \sin 7.8^\circ}{\sin 66^\circ} = \frac{18 \times 0.136}{0.9135} = 2.68 \text{ cm}.$$ 8. **Summary of solutions:** - First triangle: $A=66^\circ$, $B=40.2^\circ$, $C=73.8^\circ$, $a=18$ cm, $b=12.7$ cm, $c=18.9$ cm. - Second triangle: $A=66^\circ$, $B=7.8^\circ$, $C=106.2^\circ$, $a=18$ cm, $b=2.68$ cm, $c=18.9$ cm. Thus, there are two possible triangles given the data.