Let AB be a telegraph pole. When the sun’s angle of elevation is 45°, ∠ACB = 45° and the length of the shadow is BC. When the angle of elevation becomes 60°, ∠ADB = 60° and the shadow length is BD. So, CD = 4 meters. When the sun’s angle of elevation is ∠AEB = 30°, the shadow length is BE. From triangle ABC, with ∠ACB = 45°: \[ \tan 45^\circ = \frac{AB}{BC} \Rightarrow 1 = \frac{AB}{BC} \Rightarrow BC = AB \quad \text{---(i)} \] From triangle ABD, with ∠ADB = 60°: \[ \tan 60^\circ = \frac{AB}{BD} \Rightarrow \sqrt{3} = \frac{AB}{BD} \Rightarrow BD = \frac{AB}{\sqrt{3}} \quad \text{---(ii)} \] Now, \[ BC - BD = CD = 4 \Rightarrow AB - \frac{AB}{\sqrt{3}} = 4 \Rightarrow AB\left(1 - \frac{1}{\sqrt{3}}\right) = 4 \Rightarrow AB \times \frac{\sqrt{3} - 1}{\sqrt{3}} = 4 \Rightarrow AB = 4 \times \frac{\sqrt{3}}{\sqrt{3} - 1} \Rightarrow AB = \frac{4\sqrt{3}}{\sqrt{3} - 1} \quad \text{---(iii)} \] From triangle ABE, with ∠AEB = 30°: \[ \tan 30^\circ = \frac{AB}{BE} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{BE} \Rightarrow BE = \sqrt{3} \cdot AB \Rightarrow BE = \sqrt{3} \cdot \frac{4\sqrt{3}}{\sqrt{3} - 1} = \frac{12}{\sqrt{3} - 1} \] Rationalizing the denominator: \[ BE = \frac{12(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{12(\sqrt{3} + 1)}{3 - 1} = \frac{12(\sqrt{3} + 1)}{2} = 6(\sqrt{3} + 1) \] Therefore, when the sun’s angle of elevation is 30°, the length of the telegraph pole’s shadow is \[ 6(\sqrt{3} + 1) \text{ meters}. \]