Let the height of the shorter pillar be \(AB = x\) meters and the height of the taller pillar be \(CD = 3x\) meters. Let \(O\) be the midpoint of the line segment \(BD\) connecting the bases of the two pillars. From point \(O\), the angles of elevation to the tops of the pillars are ∠AOB = \(\theta\) and ∠COD = \(90^\circ - \theta\). Since \(O\) is the midpoint of \(BD\), \[ BO = OD = \frac{150}{2} = 75 \text{ meters} \] From triangle △ABO: \[ \tan\theta = \frac{AB}{BO} = \frac{x}{75} \quad \text{—— (i)} \] From triangle △COD: \[ \tan(90^\circ - \theta) = \frac{CD}{OD} = \frac{3x}{75} \Rightarrow \cot\theta = \frac{3x}{75} \quad \text{—— (ii)} \] Multiplying equations (i) and (ii): \[ \tan\theta \cdot \cot\theta = \frac{x}{75} \cdot \frac{3x}{75} \Rightarrow 1 = \frac{3x^2}{75 \times 75} \] Solving: \[ x^2 = \frac{75 \times 75}{3} = 25 \times 75 \Rightarrow x = 25\sqrt{3} \] Therefore, the height of the shorter pillar is \(25\sqrt{3}\) meters.