An aeroplane when flying at a height of 3000 m from the ground passes vertically above another aeroplane at an instant when the angles of elevation…

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 \(\displaystyle =3460 \mathrm{~m}\)