From an aeroplane just over a straight road, the angles of depression of two consecutive kilometre stones situated at opposite sides of the aeroplane were found to be 60° and 30° respectively. The height (in km) of the aeroplane from the road at that instant was (Given √3 = 1.732)

OC = height of plane = $$h$$

$$\angle$$OAC = $$\angle$$DOA = 60°

$$\angle$$OBC = $$\angle$$BOE = 30°

AB = 2 and let AC = $$x$$

=> BC = $$(2-x)$$

From, $$\triangle$$OAC

$$tan60^{\circ} = \frac{OC}{AC}$$

=> $$\sqrt{3} = \frac{h}{x}$$

=> $$x = \frac{h}{\sqrt{3}}$$ ------------Eqn(1)

From, $$\triangle$$OBC

$$tan30^{\circ} = \frac{OC}{BC}$$

=> $$\frac{1}{\sqrt{3}} = \frac{h}{2-x}$$

=> $$\sqrt{3}h = 2 - \frac{h}{\sqrt{3}}$$ [From eqn(1)]

=> $$\frac{3h+h}{\sqrt{3}} = 2$$

=> $$h = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$$

= $$\frac{1.732}{2}$$ = 0.866