Let AB be a chimney. From point C, the angle of elevation to the top of the chimney (point A) is ∠ACB = 30°, and after walking 50 meters forward to point D, the angle of elevation becomes ∠ADB = 60°. From right-angled triangle ABC: \[ \tan 30^\circ = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BC} \Rightarrow AB = \frac{BC}{\sqrt{3}} \quad \text{(i)} \] From right-angled triangle ABD: \[ \tan 60^\circ = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{AB}{BD} \Rightarrow AB = BD \cdot \sqrt{3} \quad \text{(ii)} \] Equating the values of AB from equations (i) and (ii): \[ \frac{BC}{\sqrt{3}} = BD \cdot \sqrt{3} \Rightarrow 3BD = BC \Rightarrow 3BD = BD + DC \Rightarrow 2BD = DC \Rightarrow BD = \frac{50}{2} = 25 \] Substituting BD into equation (ii): \[ AB = 25\sqrt{3} \] ∴ The height of the chimney is **\(25\sqrt{3}\) meters**.