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?

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

The correct answer is Option (4) → $\frac{5\sqrt{3}}{2}(\sqrt{3}+1) m$

Let the height of the pole be $h$.

Length of shadow when elevation $= 45^\circ$:

$\tan 45^\circ = \frac{h}{\text{shadow}} \Rightarrow \text{shadow} = h$

Length of shadow when elevation $= 60^\circ$:

$\tan 60^\circ = \frac{h}{\text{shadow}} \Rightarrow \text{shadow} = \frac{h}{\sqrt{3}}$

Given: shadow at $60^\circ$ is $5\text{ m}$ shorter than at $45^\circ$:

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

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

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

Rationalising:

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

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