An aeroplane flying at a height of 5000 ; m f | vedclass

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(C) Let the two aeroplanes be at points $A$ and $D$. Let the point on the ground be $C$. The height of the first aeroplane $A$ is $AB = 5000 \; m$.Let

Vedclass · Accepted Answer

$5000\left(1-\frac{1}{\sqrt{3}}\right) \; m$

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(C) Let the two aeroplanes be at points $A$ and $D$. Let the point on the ground be $C$. The height of the first aeroplane $A$ is $AB = 5000 \; m$.Let the height of the second aeroplane $D$ be $DB = h$.The distance of the point $C$ from the point $B$ (directly below the aeroplanes) is $BC = x$.In $\Delta ABC$,$	an 60^{\circ} = \frac{AB}{BC} = \frac{5000}{x}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $x = \frac{5000}{\sqrt{3}} \; m$.In $\Delta DBC$,$	an 45^{\circ} = \frac{DB}{BC} = \frac{h}{x}$.Since $	an 45^{\circ} = 1$,we have $h = x = \frac{5000}{\sqrt{3}} \; m$.The vertical distance between the aeroplanes is $AD = AB - DB = 5000 - h$.$AD = 5000 - \frac{5000}{\sqrt{3}} = 5000\left(1 - \frac{1}{\sqrt{3}}ight) \; m$.

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