Two eagles are flying in the sky at height of \(10\sqrt{3}\) m above the ground. A boy standing at a point on the ground looks towards the eagles making angles of elevation of 60º and 30º respectively. Find the distance between these two eagles.

Answer: 20 m

Let the boy stand at A, and let B and C be the positions of the two eagles, both at height \(10\sqrt{3}\) m. Let D and E be the points on the ground vertically below B and C. The eagles and the boy are in the same vertical plane, so AD and AE are the horizontal distances of the eagles from the boy.

For the eagle at 60º: \(\tan 60º = \dfrac{10\sqrt{3}}{AD}\), so AD = \(\dfrac{10\sqrt{3}}{\sqrt{3}}\) = 10 m.

For the eagle at 30º: \(\tan 30º = \dfrac{10\sqrt{3}}{AE}\), so AE = \(10\sqrt{3} \times \sqrt{3}\) = 30 m.

The eagles are on the same side of the boy, so the distance between them = AE – AD = 30 – 10 = 20 m.