1. **State the problem:** We have two right triangles $\triangle PDW$ and $\triangle WTC$. Given $\angle WPD = 30^\circ$, $WC = 6$, $WT = 3$, and $WD = 12$, we need to find the length of $PD$. 2. **Analyze the triangles:** $\triangle PDW$ is right angled at $D$, and $\triangle WTC$ is right angled at $T$. 3. **Use given lengths in $\triangle WTC$:** Since $WT = 3$ and $WC = 6$, and $\triangle WTC$ is right angled at $T$, by Pythagoras, $$WC^2 = WT^2 + TC^2 \implies 6^2 = 3^2 + TC^2 \implies 36 = 9 + TC^2 \implies TC^2 = 27 \implies TC = 3\sqrt{3}.$$ 4. **Relate $\triangle PDW$ and $\triangle WTC$:** Points $D$ and $T$ lie on the vertical line segments, and $WD = 12$, $WT = 3$, so $DW = WD - WT = 12 - 3 = 9$. 5. **Use angle $30^\circ$ in $\triangle PDW$:** In right triangle $PDW$, with right angle at $D$, $\angle WPD = 30^\circ$, side $WD = 12$ is opposite $\angle WPD$, and side $PD$ is adjacent to $\angle WPD$. Using the tangent function: $$\tan(30^\circ) = \frac{WD}{PD} \implies PD = \frac{WD}{\tan(30^\circ)}.$$ 6. **Calculate $PD$:** Recall $\tan(30^\circ) = \frac{1}{\sqrt{3}}$, so $$PD = \frac{12}{\frac{1}{\sqrt{3}}} = 12 \times \sqrt{3} = 12\sqrt{3}.$$ **Final answer:** $PD = 12\sqrt{3}$.