# An aeroplane flying with uniform speed horizontally 2 kilometer above the ground is observed at an elevation of 60°.

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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

by (21.6k points)

(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.