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An aeroplane flying with uniform speed horizontally 2 kilometer above the ground is observed at an elevation of 60°. After 10 s, if the elevation from the same point is observed to be 45°, then the distance travelled by the aeroplane is

(a) \(\frac{2(\sqrt3-1)}{\sqrt3}\) km

(b) \(2(\sqrt3+1)\) km

(c) \(\frac{\sqrt3-1}{\sqrt3}\) km

(d) None of these

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(a) \(\frac{2(\sqrt3-1)}{\sqrt3}\) km

Let O be the position of the aeroplane and A be the point on the ground. Then ∠OAM = 60°

OM = 2 km = O′N, 

where O′ is the position of the plane after 10s. 

In Δ AMO, 

tan 60° = \(\frac{OM}{AM}\) ⇒ \(\sqrt3\) = \(\frac{2}{AM}\) ⇒ AM = \(\frac{2}{\sqrt3}\) km

In Δ AO′N, tan 45° = \(\frac{O'N}{AN}\) ⇒ AN = O'N  = 2 km

∴ Distance travelled by the plane = O′O = MN

= AN – AM = 2 - \(\frac{2}{\sqrt3}\) = \(\frac{2\sqrt3-2}{\sqrt3}\) = \(\frac{2(\sqrt3-1)}{\sqrt3}\)km.

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