1. **State the problem:** We have a right triangle DEF with $\angle E = 90^\circ$, $DE = 45$, and $\angle F = 60^\circ$. We need to find the length of segment $DF$. 2. **Identify the sides and angles:** Since $\angle E = 90^\circ$, side $DF$ is the hypotenuse opposite the right angle. 3. **Use the sine function:** In a right triangle, $\sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}}$. 4. Here, $\sin(60^\circ) = \frac{DE}{DF} = \frac{45}{DF}$. 5. Solve for $DF$: $$ DF = \frac{45}{\sin(60^\circ)} $$ 6. Recall $\sin(60^\circ) = \frac{\sqrt{3}}{2}$, so $$ DF = \frac{45}{\frac{\sqrt{3}}{2}} = 45 \times \frac{2}{\sqrt{3}} = \frac{90}{\sqrt{3}}. $$ 7. Rationalize the denominator: $$ DF = \frac{90}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{90\sqrt{3}}{3} = 30\sqrt{3}. $$ **Final answer:** The length of segment $DF$ is $30\sqrt{3}$.