1. **Problem statement:** Given a vertical pole AB with point A at the top and B at the bottom on the ground, point C is 2 m to the right of B on the horizontal line. Point D lies on AB such that AD = DB, and $\angle BDC = 30^\circ$. We need to find $\angle BAC$. 2. **Understanding the problem:** Since AD = DB, point D is the midpoint of AB. Let the length of AB be $2h$, so AD = DB = $h$. 3. **Set coordinates:** Place B at the origin $(0,0)$, A at $(0,2h)$, and C at $(2,0)$. 4. **Coordinates of D:** Since D is midpoint of AB, $D = (0,h)$. 5. **Calculate vectors:** - Vector $\overrightarrow{BD} = (0,h) - (0,0) = (0,h)$ - Vector $\overrightarrow{DC} = (2,0) - (0,h) = (2,-h)$ 6. **Use the angle between vectors formula:** $$\cos \theta = \frac{\overrightarrow{BD} \cdot \overrightarrow{DC}}{|\overrightarrow{BD}||\overrightarrow{DC}|}$$ Given $\theta = 30^\circ$, so $$\cos 30^\circ = \frac{0 \cdot 2 + h \cdot (-h)}{h \sqrt{2^2 + (-h)^2}} = \frac{-h^2}{h \sqrt{4 + h^2}} = \frac{-h}{\sqrt{4 + h^2}}$$ 7. **Calculate $\cos 30^\circ = \frac{\sqrt{3}}{2}$, so:** $$\frac{\sqrt{3}}{2} = \left|\frac{-h}{\sqrt{4 + h^2}}\right| = \frac{h}{\sqrt{4 + h^2}}$$ 8. **Square both sides:** $$\left(\frac{\sqrt{3}}{2}\right)^2 = \frac{h^2}{4 + h^2} \Rightarrow \frac{3}{4} = \frac{h^2}{4 + h^2}$$ 9. **Cross multiply:** $$3(4 + h^2) = 4h^2 \Rightarrow 12 + 3h^2 = 4h^2 \Rightarrow 12 = h^2$$ 10. **Solve for $h$:** $$h = 2\sqrt{3}$$ 11. **Coordinates of A:** $(0, 2h) = (0, 4\sqrt{3})$ 12. **Calculate vectors for $\angle BAC$:** - $\overrightarrow{AB} = (0,0) - (0,4\sqrt{3}) = (0, -4\sqrt{3})$ - $\overrightarrow{AC} = (2,0) - (0,4\sqrt{3}) = (2, -4\sqrt{3})$ 13. **Find $\cos \angle BAC$:** $$\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{0 \cdot 2 + (-4\sqrt{3})(-4\sqrt{3})}{|\overrightarrow{AB}||\overrightarrow{AC}|} = \frac{48}{|\overrightarrow{AB}||\overrightarrow{AC}|}$$ 14. **Calculate magnitudes:** $$|\overrightarrow{AB}| = \sqrt{0^2 + (-4\sqrt{3})^2} = 4\sqrt{3}$$ $$|\overrightarrow{AC}| = \sqrt{2^2 + (-4\sqrt{3})^2} = \sqrt{4 + 48} = \sqrt{52} = 2\sqrt{13}$$ 15. **Calculate $\cos \angle BAC$:** $$\cos \angle BAC = \frac{48}{4\sqrt{3} \times 2\sqrt{13}} = \frac{48}{8 \sqrt{39}} = \frac{6}{\sqrt{39}}$$ 16. **Calculate $\angle BAC$:** $$\angle BAC = \cos^{-1} \left(\frac{6}{\sqrt{39}}\right) \approx 40.9^\circ$$ **Final answer:** $\boxed{40.9^\circ}$