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A person walking along a straight road observes that at two points $1 k m$ apart, the angles of elevation of a pole in front of him are $30 °$ and $75 ° .$ The height of the pole is

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$500 (\sqrt{2} - 1) m$

$250 (\sqrt{3} + 1) m$

$250 (\sqrt{3} - 1) m$

$500 (\sqrt{2} + 1) m$

answer is A.

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Detailed Solution

$h \cot 30^{\circ} - h \cot 75^{\circ} = 1000 m .$

$\begin{array}{l} \Rightarrow h = \frac{1000}{\cot 30^{\circ} - \cot 75^{\circ}} \\ = \frac{1000}{\sqrt{3} - \frac{\sqrt{3} - 1}{\sqrt{3} + 1}} = 250 (\sqrt{3} + 1) m \end{array}$

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