Question: top-left: $2$
center: In $\triangle XYZ$, $m\angle X = 55$, $m\angle Y = 39$, and $YZ = 25$. To the nearest tenth, what is $XZ$?
1. **State the problem:** We are given a triangle $XYZ$ with angles $m\angle X = 55^\circ$, $m\angle Y = 39^\circ$, and side $YZ = 25$. We need to find the length of side $XZ$ to the nearest tenth.
2. **Find the missing angle:** The sum of angles in a triangle is $180^\circ$. So,
$$m\angle Z = 180^\circ - m\angle X - m\angle Y = 180^\circ - 55^\circ - 39^\circ = 86^\circ.$$
3. **Use the Law of Sines:** The Law of Sines states:
$$\frac{XY}{\sin Z} = \frac{YZ}{\sin X} = \frac{XZ}{\sin Y}.$$
We want to find $XZ$, so we use:
$$\frac{XZ}{\sin Y} = \frac{YZ}{\sin X}.$$
4. **Plug in known values:**
$$\frac{XZ}{\sin 39^\circ} = \frac{25}{\sin 55^\circ}.$$
5. **Solve for $XZ$:**
$$XZ = \frac{25 \times \sin 39^\circ}{\sin 55^\circ}.$$
6. **Calculate sine values:**
$$\sin 39^\circ \approx 0.6293, \quad \sin 55^\circ \approx 0.8192.$$
7. **Substitute and compute:**
$$XZ \approx \frac{25 \times 0.6293}{0.8192} = \frac{15.7325}{0.8192}.$$
8. **Simplify fraction:**
$$XZ \approx 19.2.$$
**Final answer:** The length of side $XZ$ is approximately $19.2$ units.