Question: Find $AC$.
Triangle $ABC$ with $A$ at bottom-left, $B$ at bottom-right, and $C$ at upper-center-right; $AB$ is horizontal and labeled $2\sqrt{3}$, angle at $A$ is $15^\circ$, and angle at $B$ is $45^\circ$.
$AB = 2\sqrt{3}$
$\angle A = 15^\circ$
$\angle B = 45^\circ$
$AC$ is the side from $A$ to $C$, position_hint $\in \{center\}$
1. **State the problem:** We need to find the length of side $AC$ in triangle $ABC$ where $AB = 2\sqrt{3}$, $\angle A = 15^\circ$, and $\angle B = 45^\circ$.
2. **Find the missing angle:** Since the sum of angles in a triangle is $180^\circ$,
$$\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 15^\circ - 45^\circ = 120^\circ.$$
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 the sides opposite angles $A$, $B$, and $C$ respectively.
4. **Assign sides:**
- Side $AB$ is opposite $\angle C$, so $c = AB = 2\sqrt{3}$.
- Side $AC$ is opposite $\angle B$, so $b = AC$ (the side we want to find).
- Side $BC$ is opposite $\angle A$.
5. **Apply Law of Sines to find $AC$:**
$$\frac{AC}{\sin 45^\circ} = \frac{2\sqrt{3}}{\sin 120^\circ}$$
6. **Calculate sines:**
- $\sin 45^\circ = \frac{\sqrt{2}}{2}$
- $\sin 120^\circ = \sin (180^\circ - 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2}$
7. **Substitute values:**
$$\frac{AC}{\frac{\sqrt{2}}{2}} = \frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}}$$
8. **Simplify right side:**
$$\frac{2\sqrt{3}}{\frac{\sqrt{3}}{2}} = 2\sqrt{3} \times \frac{2}{\sqrt{3}} = \cancel{2} \cancel{\sqrt{3}} \times \frac{2}{\cancel{\sqrt{3}}} = 4$$
9. **Solve for $AC$:**
$$AC = 4 \times \frac{\sqrt{2}}{2} = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2}$$
**Final answer:**
$$AC = 2\sqrt{2}$$