1. **State the problem:** We have a triangle with angles 180°, 72°, and 58°, and side BC related to sine functions of 50° and 58°. We want to find the length BC using the Law of Sines. 2. **Recall the Law of Sines:** For any triangle with sides $a$, $b$, $c$ opposite angles $A$, $B$, $C$ respectively, the Law of Sines states: $$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$ 3. **Apply the Law of Sines to find BC:** Given side length opposite angle 50° is 18.4, and angle opposite BC is 58°, we write: $$\frac{BC}{\sin 50^\circ} = \frac{18.4}{\sin 58^\circ}$$ 4. **Solve for BC:** $$BC = 18.4 \times \frac{\sin 50^\circ}{\sin 58^\circ}$$ 5. **Calculate the sine values:** $$\sin 50^\circ \approx 0.7660$$ $$\sin 58^\circ \approx 0.8480$$ 6. **Substitute and compute:** $$BC = 18.4 \times \frac{0.7660}{0.8480} = 18.4 \times 0.9038 = 16.63$$ 7. **Interpretation:** The length BC is approximately 16.63 units. 8. **Regarding your question "Is that meant to be my sine?":** Yes, the sine function here is the trigonometric sine of the given angles in degrees, used to relate sides and angles in the triangle via the Law of Sines.