Suppose \( AB \) is a pole. When the sun's angle of elevation was 45°, then \( \angle ACB = 45° \) and the length of the shadow was \( BC \). When the elevation angle became 60°, then \( \angle ADB = 60° \) and the shadow length became \( BD \). ∴ \( CD = 4 \) meters.
When the sun's angle of elevation is \( \angle AEB = 30° \), the length of the shadow of the pole is \( BE \).
From triangle \( ABC \), considering \( \angle ACB \):
\(\tan 45° = \cfrac{AB}{BC} \)
Or, \( 1 = \cfrac{AB}{BC} \)
Or, \( BC = AB \) ---- (i)
From triangle \( ABD \), considering \( \angle ADB \):
\(\tan 60° = \cfrac{AB}{BD} \)
Or, \( \sqrt3 = \cfrac{AB}{BD} \)
Or, \( BD = \cfrac{AB}{\sqrt3} \) ---- (ii)
Again, \( BC - BD = CD = 4 \)
Or, \( AB - \cfrac{AB}{\sqrt3} = 4 \)
Or, \( AB (1 - \cfrac{1}{\sqrt3}) = 4 \)
Or, \( AB \times \cfrac{\sqrt3 - 1}{\sqrt3} = 4 \)
Or, \( AB = 4 \times \cfrac{\sqrt3}{\sqrt3 - 1} \)
Or, \( AB = \cfrac{4\sqrt3}{(\sqrt3 - 1)} \) ---- (iii)
From triangle \( ABE \), considering \( \angle AEB \):
\(\tan 30° = \cfrac{AB}{BE} \)
Or, \( \cfrac{1}{\sqrt3} = \cfrac{AB}{BE} \)
Or, \( BE = \sqrt3 AB \)
Or, \( BE = \sqrt3 \times \cfrac{4\sqrt3}{(\sqrt3 - 1)} = \cfrac{12}{(\sqrt3 - 1)} \)
\( = \cfrac{12(\sqrt3 + 1)}{(\sqrt3 - 1)(\sqrt3 + 1)} \)
\( = \cfrac{12(\sqrt3 + 1)}{3 - 1} \)
\( = \cfrac{12(\sqrt3 + 1)}{2} \)
\( = 6(\sqrt3 + 1) \)
\(\therefore\) When the sun's angle of elevation is 30°, the length of the shadow of the pole is \( 6(\sqrt3 + 1) \) meters.