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Two men are on the opposite sides of a tower. They measure the angles of elevation of the top of the tower as 450 and 300 respectively. If the height of the tower is 40m, then the distance between the men is

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40m

68.28m

$40 \sqrt{3} m$

109.28m

answer is D.

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Detailed Solution

From $Δ O_{1} A B, \tan 45^{0} = \frac{40}{x} \Rightarrow x = 40$ m
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From $Δ O_{2} A B, \cot 30^{0} = \frac{y}{40} \Rightarrow y = 40 \sqrt{3} m$
Distance between the men $= 40 + 40 \sqrt{3} = 109.28 m$

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