Question: In $\triangle ABC$, $m\angle A = 50$, $m\angle C = 27$, and $AB = 19$. To the nearest tenth, what is $BC$?
1. **State the problem:** We are given a triangle $ABC$ with angles $A = 50^\circ$, $C = 27^\circ$, and side $AB = 19$. We need to find the length of side $BC$ to the nearest tenth.
2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. So,
$$m\angle B = 180^\circ - m\angle A - m\angle C = 180^\circ - 50^\circ - 27^\circ = 103^\circ.$$
3. **Identify sides opposite to angles:**
- Side $BC$ is opposite angle $A$.
- Side $AC$ is opposite angle $B$.
- Side $AB$ is opposite angle $C$.
4. **Use the Law of Sines:**
$$\frac{BC}{\sin 50^\circ} = \frac{AB}{\sin 27^\circ}.$$
5. **Plug in known values:**
$$\frac{BC}{\sin 50^\circ} = \frac{19}{\sin 27^\circ}.$$
6. **Solve for $BC$:**
$$BC = \frac{19 \times \sin 50^\circ}{\sin 27^\circ}.$$
7. **Calculate sine values:**
$$\sin 50^\circ \approx 0.7660, \quad \sin 27^\circ \approx 0.4540.$$
8. **Calculate $BC$:**
$$BC = \frac{19 \times 0.7660}{0.4540} = \frac{14.554}{0.4540} \approx 32.06.$$
9. **Round to nearest tenth:**
$$BC \approx 32.1.$$
**Final answer:** The length of side $BC$ is approximately $32.1$ units.