Let AB be a pole. When the sun’s elevation angle was 45°, ∠ACB = 45° and the length of the shadow was BC. When the elevation angle becomes 60°, ∠ADB = 60° and the shadow length becomes BD. ∴ DC = 3 meters. From triangle ABC, with respect to ∠ACB: \[ \tan 45^\circ = \frac{AB}{BC} \Rightarrow 1 = \frac{AB}{BC} \Rightarrow BC = AB \quad \text{(i)} \] From triangle ABD, with respect to ∠ADB: \[ \tan 60^\circ = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{AB}{BD} \Rightarrow BD = \frac{AB}{\sqrt{3}} \quad \text{(ii)} \] Also, \[ BC - BD = CD = 3 \Rightarrow AB - \frac{AB}{\sqrt{3}} = 3 \Rightarrow AB \left(1 - \frac{1}{\sqrt{3}}\right) = 3 \Rightarrow AB \times \frac{\sqrt{3} - 1}{\sqrt{3}} = 3 \Rightarrow AB = 3 \times \frac{\sqrt{3}}{\sqrt{3} - 1} \] Now rationalizing the denominator: \[ = 3 \times \frac{\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = 3 \times \frac{3 + \sqrt{3}}{3 - 1} = 3 \times \frac{3 + 1.732}{2} = \frac{3 \times 4.732}{2} = 7.098 \] ∴ The height of the pole is 7.098 meters.