Question: Solve for angle $A$ in the diagram below.
Triangle $ABC$ with $A$ at the top, $B$ at the bottom-left, and $C$ at the right; angle at $B$ is $37^\circ$, side $BC$ is $8$ ft., and side $AC$ is $5$ ft.
1. **State the problem:** We need to find the measure of angle $A$ in triangle $ABC$ where angle $B = 37^\circ$, side $BC = 8$ ft, and side $AC = 5$ ft.
2. **Identify known elements:**
- Angle $B = 37^\circ$
- Side opposite angle $A$ is $BC = 8$ ft
- Side opposite angle $B$ is $AC = 5$ ft
3. **Formula used:** Use the Law of Sines which states:
$$\frac{\sin A}{BC} = \frac{\sin B}{AC}$$
4. **Apply the Law of Sines:**
$$\frac{\sin A}{8} = \frac{\sin 37^\circ}{5}$$
5. **Solve for $\sin A$:**
$$\sin A = 8 \times \frac{\sin 37^\circ}{5}$$
6. **Calculate $\sin 37^\circ$:**
$$\sin 37^\circ \approx 0.6018$$
7. **Substitute and simplify:**
$$\sin A = 8 \times \frac{0.6018}{5} = 8 \times 0.12036 = 0.9629$$
8. **Find angle $A$:**
$$A = \sin^{-1}(0.9629) \approx 74.4^\circ$$
9. **Check for validity:** Since $\sin A$ is less than 1, the angle is valid.
10. **Final answer:**
$$\boxed{A \approx 74.4^\circ}$$