If the angles of elevation of a balloon from two consecutive kilometre-stones along a road are 30° and 60° respectively, then the height of the balloon above the
ground will be

It is given that BC = 2 km

Let BD = x => CD = (2 - x) km

From $$\triangle$$ABD

=> $$tan 30 = \frac{AD}{BD}$$

=> $$\frac{1}{\sqrt{3}} = \frac{AD}{x}$$

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

From $$\triangle$$ADC

=> $$tan 60 = \frac{AD}{CD}$$

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

=> $$3 (2 - x) = x$$

=> $$x = \frac{3}{2}$$

Now, height of balloon above ground = $$AD = \frac{\frac{3}{2}}{\sqrt{3}}$$

= $$\frac{\sqrt{3}}{2}$$ km