1. **Problem Statement:** Given three triangles ABC with angle $A$ and sides $a$, $b$, and $c$ as described, find all possible triangles using the Ambiguous Case (SSA) of the Law of Sines. 2. **Formula and Rules:** The Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ In the Ambiguous Case, given $A$, $a$, and $b$, there can be 0, 1, or 2 possible triangles depending on the relationship between $a$, $b$, and $\sin A$. 3. **Step-by-step for each triangle:** --- ### Triangle 1: $A=45^\circ$, $a=7.07$, $b=10$ - Calculate $h = b \sin A = 10 \times \sin 45^\circ = 10 \times \frac{\sqrt{2}}{2} = 7.07$ - Since $a = h$, there is exactly one right triangle. - Find angle $B$ using Law of Sines: $$\sin B = \frac{b \sin A}{a} = \frac{10 \times \sin 45^\circ}{7.07} = \frac{7.07}{7.07} = 1$$ - So, $B = 90^\circ$. - Find angle $C = 180^\circ - A - B = 180^\circ - 45^\circ - 90^\circ = 45^\circ$. - Find side $c$ using Law of Sines: $$c = \frac{a \sin C}{\sin A} = \frac{7.07 \times \sin 45^\circ}{\sin 45^\circ} = 7.07$$ --- ### Triangle 2: $A=60^\circ$, $a=20$, $b=10$ - Calculate $h = b \sin A = 10 \times \sin 60^\circ = 10 \times \frac{\sqrt{3}}{2} = 8.66$ - Since $a > b$, and $a > h$, there is exactly one triangle. - Find angle $B$: $$\sin B = \frac{b \sin A}{a} = \frac{10 \times \sin 60^\circ}{20} = \frac{8.66}{20} = 0.433$$ - So, $B = \arcsin(0.433) \approx 25.7^\circ$. - Find angle $C = 180^\circ - A - B = 180^\circ - 60^\circ - 25.7^\circ = 94.3^\circ$. - Find side $c$: $$c = \frac{a \sin C}{\sin A} = \frac{20 \times \sin 94.3^\circ}{\sin 60^\circ} = \frac{20 \times 0.997}{0.866} \approx 23.0$$ --- ### Triangle 3: $A=30^\circ$, $a=8$, $b=10$ - Calculate $h = b \sin A = 10 \times \sin 30^\circ = 10 \times 0.5 = 5$ - Since $a > h$ and $a < b$, there are two possible triangles. - Find angle $B_1$: $$\sin B_1 = \frac{b \sin A}{a} = \frac{10 \times 0.5}{8} = 0.625$$ - So, $B_1 = \arcsin(0.625) \approx 38.7^\circ$. - Find angle $B_2 = 180^\circ - B_1 = 141.3^\circ$ (second possible angle). - For first triangle: - $C_1 = 180^\circ - A - B_1 = 180^\circ - 30^\circ - 38.7^\circ = 111.3^\circ$ - $c_1 = \frac{a \sin C_1}{\sin A} = \frac{8 \times \sin 111.3^\circ}{\sin 30^\circ} = \frac{8 \times 0.936}{0.5} = 15.0$ - For second triangle: - $C_2 = 180^\circ - A - B_2 = 180^\circ - 30^\circ - 141.3^\circ = 8.7^\circ$ - $c_2 = \frac{a \sin C_2}{\sin A} = \frac{8 \times \sin 8.7^\circ}{0.5} = \frac{8 \times 0.151}{0.5} = 2.42$ --- **Final answers:** - Triangle 1: One triangle with $B=90^\circ$, $C=45^\circ$, $c=7.07$ - Triangle 2: One triangle with $B=25.7^\circ$, $C=94.3^\circ$, $c=23.0$ - Triangle 3: Two triangles: - First: $B=38.7^\circ$, $C=111.3^\circ$, $c=15.0$ - Second: $B=141.3^\circ$, $C=8.7^\circ$, $c=2.42$