**(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.**