An aeroplane when flying at a height of $\displaystyle 3000$ m from the ground passes vertically above another aeroplane at an instant when the angles of elevation of the two planes from the same point on the ground are $\displaystyle 60$° and $\displaystyle 45$° respectively. Find the vertical distance between the aeroplanes at that instant. Also, find the distance of first plane from the point of observation. (Take $\displaystyle \sqrt{3}=1.73$ )

Let planes be located at points C and D \[\tan 45^{\circ}=1=\frac{3000-h}{x} \Rightarrow x=3000-h \] \[\tan 60^{\circ}=\sqrt{3}=\frac{3000}{x} \Rightarrow x=1000 \sqrt{3} \] Using (i) and (ii) \(\displaystyle \mathrm{h}=3000-1730\) \[\text { = } 1270 \] ∴ vertical distance between the aeroplanes \(\displaystyle =1270 \mathrm{~m}\) Also \(\displaystyle \sin 60^{\circ}=\frac{\sqrt{3}}{2}=\frac{3000}{\mathrm{z}} \Rightarrow \mathrm{z}=2000 \sqrt{3}=3460\) Distance of the first plane from the point of observation \(\displaystyle =3460 \mathrm{~m}\)