Question : An aeroplane when flying at a height of 5000 m from the ground passes vertically above another aeroplane at an instant, when the angles of elevation of the two aeroplanes from the same point on the ground are $60^{\circ}$ and $45^{\circ}$ respectively. The vertical distance between the aeroplanes at that instant is:
Option 1: $5000(\sqrt{3}-1)$m
Option 2: $5000(3-\sqrt{3})$m
Option 3: $5000(1-\frac{1}{\sqrt{3}})$m
Option 4: $4500$ m
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Correct Answer: $5000(1-\frac{1}{\sqrt{3}})$m
Solution : Let A and B be the two positions of the aeroplane. When B is vertically below and AC = 5000 metres and let BC = $y$ metre Now, in $\triangle DCA$ ⇒ $\tan 60^\circ = \frac{AC}{DC}$ ⇒ $\sqrt3 = \frac{5000}{DC}$ ⇒ $DC = \frac{5000}{\sqrt3}$ Now, in $\triangle DCB$ $\tan 45^\circ = \frac{BC}{DC}$ ⇒ $1 = \frac{BC}{\frac{5000}{\sqrt3}}$ ⇒ $BC =\frac{5000}{\sqrt3}$ $\therefore$ Vertical distance between A and B is AC – BC $=5000–{\frac{5000}{\sqrt3}}$ $=5000(1–{\frac{1}{\sqrt3}}$) Hence, the correct answer is $5000(1–{\frac{1}{\sqrt3}}$).
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