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From an aeroplane flying, vertically above a horizontal road, the angle of depression of two consecutive stones on the same side of the aeroplane are observed to be 30 degree and 60 degree, respectively. The height (in km) at which the aeroplane is flying is

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Given: Angle of depression of two consecutive stones when observed from an aeroplane are 30 degree and 60 degree.
We have to find the height at which the aeroplane is flying.
Consider,

Let, AB = h km
BC = x km and CD = 1km
Consider, right angled triangle ABC,
tan60 = AB/BC
=> \(\sqrt{3} = \frac{h}{x}\)
=> \(x = \frac{h}{\sqrt{3}}\) .....(i)
In right angled triangle ABD, we have
tan30 = AB/BD
=> \(\frac{1}{\sqrt3} = \frac{h}{x+1}\)
=> \(x+1 = \sqrt{3}h\) ....(ii)
From(i) and  (ii), we get
\(\frac{h}{\sqrt{3}} + 1 = \sqrt{3}h\)
=> \(h+\sqrt{3} = 3h\)
=> \(2h = \sqrt{3}\)
=> So, the height at which the aeroplane is flying is \(h = \frac{\sqrt{3}}{2} km\)

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