The angle of elevation of an Aeroplane from a point on the ground is…

Aeroplane fly from A to P,

Now, In ΔABC

tan 60° = \(\frac{AB}{BC }\)

⇒ \(\frac{\sqrt {3}}{1 }\) = \(\frac{AB}{BC }\)

Now, In ΔPQC

tan 30° = \(\frac{PQ}{QC }\)

⇒ \(\frac{1}{\sqrt {3}}\) = \(\frac{PQ}{QC }\)

or

⇒ \(\frac{\sqrt {3}}{3}\) = \(\frac{PQ}{QC }\)

Here PQ = AB = height = 3000m, So

⇒\(\sqrt {3}\)R = 3000 m

⇒1R = 1000 \(\sqrt {3}\) m

Here, Distance traveled by aeroplane = QB

⇒ QB = QC - BC = 3 - 1 = 2R

⇒2R = 2000 \(\sqrt {3}\) m

Therefore,

Speed = \(\frac{2000\sqrt {3}}{8.66}\) = \(\frac{2000}{5}\) = 400 ms