The shadow of the pole standing on a level surface is found to be 5 m shorter when the sun's elevation is 60° than when it is 45°. What is the height of the pole?

Let $h$ = height of the pole, $x$ = shadow length when sun is at 60°, and $y$ = shadow length when sun is at 45°.

Given: $y - x = 5$ (shadow is 5m shorter at 60° than at 45°)

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When sun's elevation is 45°:

$\tan(45°) = \frac{h}{y}$

Since $\tan(45°) = 1$:

$y = h$

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When sun's elevation is 60°:

$\tan(60°) = \frac{h}{x}$

Since $\tan(60°) = \sqrt{3}$:

$x = \frac{h}{\sqrt{3}}$

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Using the shadow difference:

$y - x = 5$

$h - \frac{h}{\sqrt{3}} = 5$

$h\left(1 - \frac{1}{\sqrt{3}}\right) = 5$

$h\left(\frac{\sqrt{3} - 1}{\sqrt{3}}\right) = 5$

$h(\sqrt{3} - 1) = 5\sqrt{3}$

$h = \frac{5\sqrt{3}}{\sqrt{3} - 1}$

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Rationalizing by multiplying numerator and denominator by $(\sqrt{3} + 1)$:

$h = \frac{5\sqrt{3}}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$h = \frac{5\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2}$

$h = \frac{5\sqrt{3}(\sqrt{3} + 1)}{3 - 1}$

$h = \frac{5\sqrt{3}(\sqrt{3} + 1)}{2}$

$h = \frac{5\sqrt{3}}{2}(\sqrt{3} + 1)$

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The answer is option 4: $\frac{5\sqrt{3}}{2}(\sqrt{3} + 1)$ m