From the top of a cliff 300 meters high, the top of a tower was observed at an angle fo depression 30^oand from the foot of the tower the top of the cliff was observed at a angle of elevation 45^o. The height of the tower is 100 $λ$ then $λ$ =

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Detailed Solution

$Let AB be the cliff with AB = 300 m Let DE be the tower with height h Let BE = xm In right angles △ ABE, \begin{array}{l} \frac{AB}{BE} = \tan 45^{\circ} \\ \Rightarrow \frac{300}{x} = 1 \\ \Rightarrow x = 300 m In right angled △ ACD \begin{array}{l} \frac{AC}{CD} = \tan 30^{\circ} \\ \Rightarrow \frac{300 - h}{x} = \frac{1}{\sqrt{3}} \\ \Rightarrow \frac{300 - h}{300} = \frac{1}{\sqrt{3}} \\ \Rightarrow 300 \sqrt{3} - \sqrt{3} h = 300 \\ \Rightarrow 300 \sqrt{3} - 300 = \sqrt{3} h \\ \Rightarrow h = \frac{300 (\sqrt{3} - 1)}{\sqrt{3}} = 100 (3 - \sqrt{3}) \end{array} \end{array}$