Question: What is the length of side $BC$ in the diagram below?
Triangle $ABC$ with $A$ at the top, $B$ at the left, and $C$ at the lower right; side $AB$ is labeled $10$ km, angle at $A$ is $64^\circ$, angle at $B$ is $74^\circ$, and side $BC$ is the bottom slanted side from $B$ to $C$; position_hint = center
1. **State the problem:** We need to find the length of side $BC$ in triangle $ABC$ where $AB=10$ km, $\angle A=64^\circ$, and $\angle B=74^\circ$.
2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$.
$$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 64^\circ - 74^\circ = 42^\circ$$
3. **Use the Law of Sines:** The Law of Sines states:
$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$$
where $a$, $b$, and $c$ are sides opposite angles $A$, $B$, and $C$ respectively.
4. **Identify sides and angles:** Side $AB=10$ km is opposite angle $C=42^\circ$.
We want side $BC$, which is opposite angle $A=64^\circ$.
5. **Set up the ratio:**
$$\frac{BC}{\sin 64^\circ} = \frac{10}{\sin 42^\circ}$$
6. **Solve for $BC$:**
$$BC = \frac{10 \times \sin 64^\circ}{\sin 42^\circ}$$
7. **Calculate the sines:**
$\sin 64^\circ \approx 0.8988$
$\sin 42^\circ \approx 0.6691$
8. **Calculate $BC$:**
$$BC = \frac{10 \times 0.8988}{0.6691} \approx \frac{8.988}{0.6691} \approx 13.43$$
9. **Final answer:** The length of side $BC$ is approximately $13.43$ km.