From a point on the level ground the angle of elevation of the top of a pole is 30° on moving 20 meters nearer, the angle of elevation is 45° then the height of the pole in meters is

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$10 (\sqrt{3} - 1)$

$10 (\sqrt{3} + 1)$

answer is B.

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Detailed Solution

$\begin{array}{l} \tan 45^{\circ} = \frac{h}{x} \Rightarrow h = x \\ \tan 30^{\circ} = \frac{h}{x + 20} \\ (x + 20) \frac{1}{\sqrt{3}} = h \\ \Rightarrow h + 20 = \sqrt{3} h \\ h = \frac{20}{\sqrt{3} - 1} \end{array}$
$\begin{array}{l} h = \frac{20}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\ = \frac{20 (\sqrt{3} + 1)}{2} = 10 (\sqrt{3} + 1) \end{array}$

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