Question 137

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

Solution

BC = 2 km

Let BD = x => DC = (2-x)

In $$\triangle$$ABD

=> $$tan \angle ABD = \frac{AD}{BD}$$

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

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

Now, In $$\triangle$$ADC

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

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

=> $$2\sqrt{3} - 3AD = AD$$

=> $$AD = \frac{\sqrt{3}}{2}$$

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