(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.