1. **Problem Statement:** Given a triangle with two sides and a non-included angle (SSA), use the Law of Sines to find the unknown sides or angles. 2. **Law of Sines Formula:** $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ where $a$, $b$, and $c$ are sides opposite angles $A$, $B$, and $C$ respectively. 3. **Important Rule:** SSA can produce 0, 1, or 2 possible triangles (ambiguous case). 4. **Step-by-step:** - Identify the known side and angle pairs. - Use the Law of Sines to find the unknown angle: $$\sin B = \frac{b \sin A}{a}$$ - Calculate $B = \arcsin\left(\frac{b \sin A}{a}\right)$. - Check for the ambiguous case: if $B$ is acute, then $180^\circ - B$ might also be a solution. - Find the third angle $C = 180^\circ - A - B$. - Use Law of Sines again to find the remaining side. 5. **Example:** Suppose $a=7$, $A=30^\circ$, $b=10$. $$\sin B = \frac{10 \times \sin 30^\circ}{7} = \frac{10 \times 0.5}{7} = \frac{5}{7} \approx 0.7143$$ $$B = \arcsin(0.7143) \approx 45.57^\circ$$ Possible second solution: $$B' = 180^\circ - 45.57^\circ = 134.43^\circ$$ 6. **Check if $C$ is positive:** $$C = 180^\circ - 30^\circ - 45.57^\circ = 104.43^\circ$$ (valid) $$C' = 180^\circ - 30^\circ - 134.43^\circ = 15.57^\circ$$ (also valid) 7. **Find side $c$ for both cases using Law of Sines:** $$c = \frac{a \sin C}{\sin A} = \frac{7 \times \sin 104.43^\circ}{\sin 30^\circ} = \frac{7 \times 0.9703}{0.5} = 13.58$$ $$c' = \frac{7 \times \sin 15.57^\circ}{0.5} = \frac{7 \times 0.268}{0.5} = 3.75$$ **Summary:** SSA can yield two triangles with angles $(30^\circ, 45.57^\circ, 104.43^\circ)$ and sides $(7, 10, 13.58)$ or $(30^\circ, 134.43^\circ, 15.57^\circ)$ and sides $(7, 10, 3.75)$.